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.util.*;

public class Main {
	static int mod = 1000000007;
  static int size = 200000;
	static long[] fac = new long[size];
	static long[] finv = new long[size];
	static long[] inv = new long[size];
	static int INF = Integer.MAX_VALUE;

 	public static void main(String[] args){
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] c = new int[n+1];
		for(int i = 1; i <= n; i++){
			c[i] = scanner.nextInt();
		}
		String[] s = new String[n+1];
		String[] t = new String[n+1];
		for(int i = 1; i <= n; i++){
			s[i] = scanner.next();
			StringBuffer sb = new StringBuffer(s[i]);
			t[i] = sb.reverse().toString();
		}
		//dp[i][j]:= i番目まで見て、i番目の文字列をj=0そのまま、j=1反転使用した時に払う
		//時のコストの総和
		long[][] dp = new long[n+1][2];
		for(int i = 1; i <= n; i++){
			for(int j = 0; j < 2; j++){
				dp[i][j] = Long.MAX_VALUE/2;
			}
		}
		dp[1][0] = 0;
		dp[1][1] = c[1];
		for(int i = 2; i <= n; i++){
			if(s[i].compareTo(s[i-1]) >= 0){
				dp[i][0] = dp[i-1][0];
			}
			if(s[i].compareTo(t[i-1]) >= 0){
				dp[i][0] = Math.min(dp[i-1][1], dp[i][0]);
			}
			if(t[i].compareTo(s[i-1]) >= 0){
				dp[i][1] = dp[i-1][0]+c[i];
			}
			if(t[i].compareTo(t[i-1]) >= 0){
				dp[i][1] = Math.min(dp[i][1], dp[i-1][1]+c[i]);
			}
		}
		if(Math.min(dp[n][0],dp[n][1]) == Long.MAX_VALUE/2){
			System.out.println(-1);
			return;
		}
		System.out.println(Math.min(dp[n][0],dp[n][1]));
	}
	public static class Edge{
		int to;
		Edge(int to){
			this.to = to;
		}
	}
	public static void swap(long a, long b){
		long tmp = 0;
		if(a > b){
			tmp = a;
			a = b;
			b = tmp;
		}
	}
  static class Pair implements Comparable<Pair>{
    int first, second;
    Pair(int a, int b){
        first = a;
        second = b;
    }
    @Override
    public boolean equals(Object o){
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return first == p.first && second == p.second;
    }
    @Override
    public int compareTo(Pair p){
        //return first == p.first ? second - p.second : first - p.first; //firstで昇順にソート
        //return (first == p.first ? second - p.second : first - p.first) * -1; //firstで降順にソート
        return second == p.second ? first - p.first : second - p.second;//secondで昇順にソート
        //return (second == p.second ? first - p.first : second - p.second)*-1;//secondで降順にソート
    }
  }

  //繰り返し二乗法
  public static long pow(long x, long n){
    long ans = 1;
    while(n > 0){
      if((n & 1) == 1){
        ans = ans * x;
        ans %= mod;
      }
      x = x * x % mod;
      n >>= 1;
    }
    return ans;
  }

	public static long div(long x, long y){
		return (x*pow(y, mod-2))%mod;
	}

  //fac, inv, finvテーブルの初期化、これ使う場合はinitComb()で初期化必要
	public static  void initComb(){
		fac[0] = finv[0] = inv[0] = fac[1] = finv[1] = inv[1] = 1;
		for (int i = 2; i < size; ++i) {
			fac[i] = fac[i - 1] * i % mod;
			inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
			finv[i] = finv[i - 1] * inv[i] % mod;
		}
	}

	//nCk % mod
	public static long comb(int n, int k){
		return fac[n] * finv[k] % mod * finv[n - k] % mod;
	}

	//n! % mod
	public static long fact(int n){
		return fac[n];
	}

	//(n!)^-1 with % mod
	public static long finv(int n){
		return finv[n];
	}

  static class UnionFind {
    int[] parent;
    public UnionFind(int size) {
      parent = new int[size];
      Arrays.fill(parent, -1);
    }
    public boolean unite(int x, int y) {
      x = root(x);
      y = root(y);
      if (x != y) {
        if (parent[y] < parent[x]) {
          int tmp = y;
          y = x;
          x = tmp;
        }
        parent[x] += parent[y];
        parent[y] = x;
        return true;
      }
      return false;
    }
    public boolean same(int x, int y) {
      return root(x) == root(y);
    }
    public int root(int x) {
      return parent[x] < 0 ? x : (parent[x] = root(parent[x]));
    }
    public int size(int x) {
      return -parent[root(x)];
    }
  }
	public static int upperBound(long[] array, long value) {
			 int low = 0;
			 int high = array.length;
			 int mid;
			 while( low < high ) {
					 mid = ((high - low) >>> 1) + low; // (high + low) / 2
					 if( array[mid] <= value ) {
							 low = mid + 1;
					 } else {
							 high = mid;
					 }
			 }
			 return low;
	 }
	 public static final int lowerBound(final long[] arr, final long value) {
    int low = 0;
    int high = arr.length;
    int mid;
    while (low < high) {
        mid = ((high - low) >>> 1) + low;    //(low + high) / 2 (オーバーフロー対策)
        if (arr[mid] < value) {
            low = mid + 1;
        } else {
            high = mid;
        }
    }
    return low;
}
  //n,mの最大公約数
  public static long gcd(long n, long m){
    if(m > n) return gcd(m,n);
    if(m == 0) return n;
    return gcd(m, n%m);
  }
	//3要素のソート
	private class Pair2 implements Comparable<Pair2> {
		String s;
		int p;
		int index;
		public Pair2(String s, int p, int index) {
				this.s = s;
				this.p = p;
				this.index = index;
		}

		public int compareTo(Pair2 other) {
				if (s.equals(other.s)) {
						return other.p - this.p;
				}
				return this.s.compareTo(other.s);
		}
	}
	//c -> intに変換
public static long c2i(char c){
	if('A' <= c && c <= 'Z'){
		return c - 'A';
	}else{
		return c - 'a' + 26;
	}
}
// int -> toに変換
public static char i2c(long i){
	if(0 <= i && i < 26){
			return (char)(i + 'A');
	}else{
			return (char)(i + 'a' - 26);
	}
}

}
