Vasiliy is fond of solving different tasks. Today he found one he wasn't able to solve himself, so he asks you to help.

Vasiliy is given n strings consisting of lowercase English letters. He wants them to be sorted in lexicographical order (as in the dictionary), but he is not allowed to swap any of them. The only operation he is allowed to do is to reverse any of them (first character becomes last, second becomes one before last and so on).

To reverse the i-th string Vasiliy has to spent ci units of energy. He is interested in the minimum amount of energy he has to spent in order to have strings sorted in lexicographical order.

String A is lexicographically smaller than string B if it is shorter than B (|A| < |B|) and is its prefix, or if none of them is a prefix of the other and at the first position where they differ character in A is smaller than the character in B.

For the purpose of this problem, two equal strings nearby do not break the condition of sequence being sorted lexicographically.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings.

The second line contains n integers ci (0 ≤ ci ≤ 109), the i-th of them is equal to the amount of energy Vasiliy has to spent in order to reverse the i-th string. 

Then follow n lines, each containing a string consisting of lowercase English letters. The total length of these strings doesn't exceed 100 000.

Output

If it is impossible to reverse some of the strings such that they will be located in lexicographical order, print  - 1. Otherwise, print the minimum total amount of energy Vasiliy has to spent.

Examples

Input

2
1 2
ba
ac


Output

1


Input

3
1 3 1
aa
ba
ac


Output

1


Input

2
5 5
bbb
aaa


Output

-1


Input

2
3 3
aaa
aa


Output

-1

Note

In the second sample one has to reverse string 2 or string 3. To amount of energy required to reverse the string 3 is smaller.

In the third sample, both strings do not change after reverse and they go in the wrong order, so the answer is  - 1.

In the fourth sample, both strings consists of characters 'a' only, but in the sorted order string "aa" should go before string "aaa", thus the answer is  - 1.






import java.io.*;
import java.util.*;
import java.math.BigInteger; 
import java.lang.*;





public class Main {
    
    
    
    
   static class sort implements Comparator<int[]>
   {
        public int compare(int[] a,int[] b)
        {
            
            return -b[0]+a[0];
        }
   }
   static class sort1 implements Comparator<int[]>
   {
        public int compare(int[] a,int[] b)
        {
            
          return  b[0]-a[0];
        }
   }
   static class sort5 implements Comparator<long[]>
   {
        public int compare(long[] a,long[] b)
        {
            
           double d1,d2;
           d1 = (a[0]+0.00)/(a[1]+0.00);
           d2 = (b[0]+0.00)/(b[1]+0.00);
           //int v = Double.comapre(d2,d1);
           double eph = 0.000000000000001;
           double d = (d2-d1);
          // pr(d1+" "+d2);
           if(d>eph) return 1;
           else if(d<eph) return -1;
           if(a[1]>b[1]) return 1;
           else if(b[1]>a[1]) return -1;
           return 0;
        }
   }
   static class sort11 implements Comparator<double[]>
   {
        public int compare(double[] a,double[] b)
        {
            //if(a[0] == b[0]) return a[1]-b[1];
           if(a[1] < b[1]) return -1;
           else if(a[1] > b[1]) return 1;
           return 0;
        }
   }
   public static String[] F(BufferedReader bf) throws Exception
    {
        return (bf.readLine().split(" "));
    }
    
   public static void pr(PrintWriter out,Object o)
   {
      
    out.println(o.toString());//out.flush();
    
   }
   public static void prW(PrintWriter out,Object o)
   {
      
    out.print(o.toString());//out.flush();
    
   }
   public static int intIn(String st)
   {
      return Integer.parseInt(st);
   }
   
   
    
    public static void pr(Object o)
    {
        System.out.println(o.toString());
    }
    public static void prW(Object o)
    {
        System.out.print(o.toString());
    }
    
   
    public static int inInt(String s)
    {
        return Integer.parseInt(s);
    }
    public static long in(String s)
    {
        return Long.parseLong(s);
    }
 
 
    static int[] toIntArray(String[] m) 
    { 
        int[] p=new int[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= inInt(m[o]);
        }
        return p;
    }
    static double[] toDArray(String[] m) 
    { 
        double[] p=new double[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= Double.parseDouble(m[o]);
        }
        return p;
    }
    static long[] toLArray(String[] m) 
    { 
       long[] p=new long[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= in(m[o]);
        }
        return p;
    }
    
    static int[][] di={{0,1},{1,0},{0,-1},{-1,0}};
    static int[] dir = {4,3,2,-4,-3,-2};
    
     static long gcd(long a, long b)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b); 
        }
        static long Gp(long l,long b,long lcm)
        {
            long c1,c2;
                c1 = (l/lcm);
                long sum=0l;
                sum += (c1-1l)*(b);
                long las = (l/lcm)*lcm;
                sum += Math.min(l-las+1l,b);
                return sum;
        }
        static long pow(long x, long y, long p)
      {
        if(y == 0) return 1l;
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0l; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
    
    public static long F(int a,int b)
    {
         return 462161773423l*(a+0l) + 4549834777463l*(b+0l);
    }
    static int nCr(int n, int r, int p)
    {
        if (r > n - r)
            r = n - r;
 
        // The array C is going to store last
        // row of pascal triangle at the end.
        // And last entry of last row is nCr
        int C[] = new int[r + 1];
 
        C[0] = 1; // Top row of Pascal Triangle
 
        // One by constructs remaining rows of Pascal
        // Triangle from top to bottom
        for (int i = 1; i <= n; i++) {
 
            // Fill entries of current row using previous
            // row values
            for (int j = Math.min(i, r); j > 0; j--)
 
                // nCj = (n-1)Cj + (n-1)C(j-1);
                C[j] = (C[j] + C[j - 1]) % p;
        }
        return C[r];
    }
    // F(o,-1,tm,vis,par,o,mat);
   public static int log(int x)
   {
       boolean bol = true;

       if(((x-1)&x) == 0) bol=false;

       int c=0;
       while(x > 0)
       {
        x=(x/2);
        c++;
       }

       if(!bol) c--;
       return c;

   }

   public static int F(int[] arr,int j,int x,int n)
   {

    if(j>=n) return 0;
    if(j == (n-1)) return 0;
    boolean bt = false;
       int i = j;

       while(i<n && arr[i]<= x)
       {
          i++;
       }
       int k = j;
       //if(i == n) return 0;
       i--;
       while(k<i)
       {
          if(arr[k]>arr[k+1]) return -10000000;
          k++;
       }
       k = i+1;
       while(k<n)
       {
            if(arr[k] < x ) bt=true;
            k++;
       }

       k = j;
       boolean bol = true;
       while(k<(n-1))
       {
          if(arr[k]>arr[k+1]) 
          {
            bol=false;break;
          }

          k = k+1;
       }
       if(bol) return 0;

       int tem = arr[i+1];
       arr[i+1] = x;
       //x=tem;
       int sum;
       if(!bt)
        sum =  1 + F(arr,i+2,tem,n);
        else sum=-10000000;
     // System.out.println(sum+" "+j+" "+x);
       return sum;

   }
   static class sor implements Comparator<long[]>
   {
      public int compare(long[] a,long[] b)
      {
         if(a[0] == b[0])
         {
            if(a[1] < b[1]) return -1;
            return 1;
         }
         if(a[0] > b[0]) return 1;
         return -1;
      }
   }

   public static void F(int i,int[] vis,int v,List<Set<Integer>> list,TreeMap<Integer,Integer> tm,int[] arr)
   {
        vis[i] = v;
        if(tm.containsKey(arr[i])) tm.put(arr[i],tm.get(arr[i])+1);
        else tm.put(arr[i],1);
        for(int r : list.get(i))
        {
            if(vis[r] == 0)
            {
                F(r,vis,v,list,tm,arr);
            }
        }
   }
    static long ncr(int n, int r)
    {
 
        // p holds the value of n*(n-1)*(n-2)...,
        // k holds the value of r*(r-1)...
        long p = 1, k = 1;
        if(n<r) return 0l;
 
        // C(n, r) == C(n, n-r),
        // choosing the smaller value
        if (n - r < r) {
            r = n - r;
        }
 
        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;
 
                // gcd of p, k
                long m = __gcd(p, k);
 
                // dividing by gcd, to simplify
                // product division by their gcd
                // saves from the overflow
                p /= m;
                k /= m;
 
                n--;
                r--;
            }
 
            // k should be simplified to 1
            // as C(n, r) is a natural number
            // (denominator should be 1 ) .
        }
        else {
            p = 1;
        }
 
        // if our approach is correct p = ans and k =1
        return p;
    }
 
    static long __gcd(long n1, long n2)
    {
        long gcd = 1;
 
        for (int i = 1; i <= n1 && i <= n2; ++i) {
            // Checks if i is factor of both integers
            if (n1 % i == 0 && n2 % i == 0) {
                gcd = i;
            }
        }
        return gcd;
    }
    public static boolean F(double x,double[] arr,double k)
    {
         double eph = 0.0000001;
         double ans=0.00;
         for(double d : arr)
         {
                int c = Double.compare(d,x);
                if(c<0)
                {
                        ans += (x-d);
                }
                else if(c>0)
                {
                        ans -= ((d-x)*(1.00-k));
                }
         }

         return  (ans <= eph); 
    }
    public static boolean F(int x,int b,boolean[] ok,List<List<int[]>> p)
    {

        boolean bol = false;
        for(int[] y : p.get(x))
        {
            int a1,a2;
            a1=y[0];a2=y[1];

            if(a1== b) continue;
            if(a2 == 2) bol=true;
            boolean bt = F(a1,x,ok,p);
            bol = (bol || bt);
        }

        ok[x]=bol;
        return bol;
    }
    public static int F(String s1,String s2,int a,int b)
    {
       int i,j;
       int n1,n2;
       n1=s1.length();n2=s2.length();
       if(a==1) s1=((new StringBuilder(s1)).reverse().toString());
       if(b==1) s2 =((new StringBuilder(s2)).reverse().toString());

       return s1.compareTo(s2);

       
    }
    public static long F(int i,int k,Long[][] dp,long[] arr,String[] list)
    {
      if(i>= arr.length) return 0l;
      if(dp[i][k] != null)  return dp[i][k];


      long s1,s2;
      s1 = 100000000000001l;s2=s1;
      String st1,st2;
      st1=list[i-1];st2=list[i];
      if(k==0)
      {
            if(F(st1,st2,0,0) <= 0)
            {
                  s1 = F(i+1,0,dp,arr,list);
            }
            if(F(st1,st2,0,1) <= 0)
            {
                  s1 = Math.min(s1,arr[i]+F(i+1,1,dp,arr,list));
            }
      }
      else
      {
            if(F(st1,st2,1,0) <= 0)
            {
                  s1 = Math.min(s1,F(i+1,0,dp,arr,list));
            
            }
            if(F(st1,st2,1,1) <= 0)
            {
                  s1 = Math.min(s1,arr[i]+F(i+1,1,dp,arr,list));
            }
      }

      dp[i][k] = s1;return s1;
        
        

        
    }
    public static void main (String[] args) throws Exception {
        
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);;;
        
        //int[] map=new int[1000001];
      
       int yy=1;//inInt(bf.readLine());
       


        
        for(int w=0;w<yy;w++)
        {
            // String str = bf.readLine();
          out.flush();
         
            String[] xlp = bf.readLine().split(" ");;;;;;
            //String st = bf.readLine();
         int n;//boolean bol=false;
           long m;//long a,b,c;
           int ioo;
           long  k;//pr(out,"vbc");
         n=inInt(xlp[0]);//m=in(xlp[1]);//k=in(xlp[2]);//t=inInt(xlp[3]);

         long[] arr = toLArray(F(bf));
         String[] str=new String[n];
         for(int o=0;o<n;o++)
         {
            str[o] = bf.readLine();  
         }
         Long[][] h = new Long[n][2];
         long ans = Math.min(F(1,0,h,arr,str),arr[0]+F(1,1,h,arr,str));
         if(ans >= (100000000000001l)) pr(out,-1);
         else pr(out,ans);
         
         



         
         
   }
    out.close();
   bf.close();
}}


/*
11
3 9 9 6 4 3 6 4 9 6 0


Kickstart
 String rp;
            rp = "Case #"+(w+1)+": "+(n-ans)+" ";
static int[][] dir={{0,1},{1,0},{-1,0},{0,-1}};

static class SegmentTreeRMQ 
    { 
    int st[]; 
    int minVal(int x, int y) { 
        return (x > y) ? x : y; 
    } 
  
    
    int getMid(int s, int e) { 
        return s + (e - s) / 2; 
    } 
  
    
    int RMQUtil(int ss, int se, int qs, int qe, int index) 
    { 
        
        if (qs <= ss && qe >= se) 
            return st[index]; 
  
        // If segment of this node is outside the given range 
        if (se < qs || ss > qe) 
            return Integer.MIN_VALUE; 
  
        // If a part of this segment overlaps with the given range 
        int mid = getMid(ss, se); 
        return minVal(RMQUtil(ss, mid, qs, qe, 2 * index + 1), 
                RMQUtil(mid + 1, se, qs, qe, 2 * index + 2)); 
    } 
  
    // Return minimum of elements in range from index qs (query 
    // start) to qe (query end).  It mainly uses RMQUtil() 
    int RMQ(int n, int qs, int qe) 
    { 
        // Check for erroneous input values 
        
  
        return RMQUtil(0, n - 1, qs, qe, 0); 
    } 
  
    // A recursive function that constructs Segment Tree for 
    // array[ss..se]. si is index of current node in segment tree st 
    int constructSTUtil(int arr[], int ss, int se, int si) 
    { 
        // If there is one element in array, store it in current 
        //  node of segment tree and return 
        if (ss == se) { 
            st[si] = arr[ss]; 
            return arr[ss]; 
        } 
  
        // If there are more than one elements, then recur for left and 
        // right subtrees and store the minimum of two values in this node 
        int mid = getMid(ss, se); 
        st[si] = minVal(constructSTUtil(arr, ss, mid, si * 2 + 1), 
                constructSTUtil(arr, mid + 1, se, si * 2 + 2)); 
        return st[si]; 
    } 
  
    
    void con(int arr[]) 
    { 
        // Allocate memory for segment tree 
  
        //Height of segment tree 
        int n = (arr.length);
        int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); 
  
        //Maximum size of segment tree 
        int max_size = 2 * (int) Math.pow(2, x) - 1; 
        st = new int[max_size]; // allocate memory 
  
        // Fill the allocated memory st 
        constructSTUtil(arr, 0, n - 1, 0); 
    }
    }
     static class DSU {
    
    int[] p;int[] sz;int op;int c;;
    int[] last;
    public void G(int n)
    {
        last=new int[n];
        p=new int[n];
        sz=new int[n];c=n;
        op=n;
        for(int h=0;h<n;h++)
        {
            sz[h]=1;p[h]=h;
            last[h]=h;
        }
    }
    public int find(int x)
    {
        int y=x;
        while(x!=p[x]) x=p[x];
        while(y!=p[y])
        {
            int tem=p[y];
            p[y]=x;y=tem;
        }
        return p[y];
    }
    public void union(int a,int b)
    {
        int x,y;
        x=find(a);y=find(b);
        if(x==y) return;
        if(sz[x]>sz[y])
        {
            p[y] = x;
            sz[x]+=sz[y];
            last[x]=Math.max(last[x],last[y]);
        }
        else
        {
            p[x]=y;sz[y]+=sz[x];
            last[y]=Math.max(last[y],last[x]);
        }
        c--;
        
    }}

    static long pow(long x, long y, long p)
      {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0l; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
      static long gcd(long a, long b)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b); 
        }
        static int gcd(int a, int b,int o)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b,o); 
        }

Geometric median

public static double F(double[] x,double[] w)
    {
        double d1,d2;
        double S=0.00;
        for(double dp : w) S += dp;
        int k = 0;
        double sum = S - w[0]; // sum is the total weight of all `x[i] > x[k]`

        while(sum > S/2)
        {
            ++k;
            sum -= w[k];
        }
        d1=x[k];
        return d1;
        k = w.length-1;
       sum = S - w[k]; // sum is the total weight of all `x[i] > x[k]`

        while(sum > S/2)
        {
            --k;
            sum -= w[k];
        }
        d2=x[k];
        return new double[]{d1,d2};
    }
*/



























