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.math.*;
import java.util.*;

/**
 *
 * @author Saju
 *
 */

public class Main {

    private static int dx[] = { 1, 0, -1, 0 };
    private static int dy[] = { 0, -1, 0, 1 };

    private static final long INF = (long) Math.pow(10, 16);
    private static final int INT_INF = Integer.MAX_VALUE;
    private static final long NEG_INF = Long.MIN_VALUE;
    private static final int NEG_INT_INF = Integer.MIN_VALUE;
    private static final double EPSILON = 1e-10;

    private static final long MAX = (long) 1e12;
    
    private static final long MOD = 100000007;

    private static final int MAXN = 300005;
    private static final int MAXA = 2000007;
    private static final int MAXLOG = 22;
    private static final double PI = Math.acos(-1);
    
	public static void main(String[] args) throws IOException {

		InputReader in = new InputReader(System.in);
//		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

//         InputReader in = new InputReader(new FileInputStream("src/test.in"));
//         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("src/test.out")));

		
		
/*



*/		
		
		
		int n = in.nextInt();
		int[] cost = in.nextIntArr(n);
		
		String[] str = new String[n];
		String[] rev = new String[n];
		for(int i = 0; i < n; i++) {
			String s = in.next();
			StringBuilder sb = new StringBuilder();
			sb.append(s);
			str[i] = new String(s);
			rev[i] = new String(sb.reverse().toString());
		}
		long dp[][] = new long[n][2];
		for (int i = 0; i < n; i++) {
			dp[i][0] = INF;
			dp[i][1] = INF;
		}
		dp[0][0] = 0;
		dp[0][1] = cost[0];

		for (int i = 1; i < n; i++) {
			if (str[i - 1].compareTo(str[i]) <= 0) {
				dp[i][0] = min(dp[i][0], dp[i - 1][0]);
			}
			if (str[i - 1].compareTo(rev[i].toString()) <= 0) {
				dp[i][1] = min(dp[i][1], dp[i - 1][0] + cost[i]);
			}
			if (rev[i - 1].compareTo(str[i]) <= 0) {
				dp[i][0] = min(dp[i][0], dp[i - 1][1]);
			}
			if (rev[i - 1].compareTo(rev[i]) <= 0) {
				dp[i][1] = min(dp[i][1], dp[i - 1][1] + cost[i]);
			}
		}
		long ans = min(dp[n - 1][0], dp[n - 1][1]);
		if (ans >= INF) {
			out.println("-1");
		} else {
			out.println(ans);
		}
			
		in.close();
		out.flush();
		out.close();
		System.exit(0);
	}	
	
	
	/*
	 * return the number of elements in list that are less than or equal to the val
	 */
	private static long upperBound(List<Long> list, long val) {
		int start = 0;
		int len = list.size();
		int end = len - 1;
		int mid = 0;

		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			long v = list.get(mid);
			if (v == val) {
				start = mid;
				while(start < end) {
					mid = (start + end) / 2;
					if(list.get(mid) == val) {
						if(mid + 1 < len && list.get(mid + 1) == val) {
							start = mid + 1;
						}
						else {
							return mid + 1;
						}
					}
					else {
						end = mid - 1;
					}
				}
				return start + 1;
			}
			if (v > val) {
				end = mid - 1;
			} else {
				start = mid + 1;
			}
		}
		if (list.get(mid) < val) {
			return mid + 1;
		}
		return mid;
	}


	private static boolean isPalindrome(String str) {
		StringBuilder sb = new StringBuilder();
		sb.append(str);
		String str1 = sb.reverse().toString();
		return str.equals(str1);
	}

	private static String getBinaryStr(long n, int j) {
		String str = Long.toBinaryString(n);
		int k = str.length();
		for (int i = 1; i <= j - k; i++) {
			str = "0" + str;
		}

		return str;
	}
	
	private static long modInverse(long r) {
		return bigMod(r, MOD - 2, MOD);
	}
	
	private static long bigMod(long n, long k, long m) {

        long ans = 1;
        while (k > 0) {
            if ((k & 1) == 1) {
                ans = (ans * n) % m;
            }
            n = (n * n) % m;
            k >>= 1;
        }
        return ans;
    }
    
	private static long ceil(long n, long x) {
		long div = n / x;
        if(div * x != n) {
            div++;
        }
        return div;
	}
	
    private static int ceil(int n, int x) {
        int div = n / x;
        if(div * x != n) {
            div++;
        }
        return div;
    }

    
    private static int abs(int x) {
        if (x < 0) {
            return -x;
        }
        return x;
    }
    
    private static double abs(double x) {
        if (x < 0) {
            return -x;
        }
        return x;
    }

    private static long abs(long x) {
        if(x < 0) {
            return -x;
        }
        return x;
    }

    private static int lcm(int a, int b) {
        return (a * b) / gcd(a, b);
    }

    private static int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
    
    private static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    private static int log(long x, int base) {
        return (int) (Math.log(x) / Math.log(base));
    }

    private static long min(long a, long b) {
        if (a < b) {
            return a;
        }
        return b;
    }

    private static int min(int a, int b) {
        if (a < b) {
            return a;
        }
        return b;
    }

    private static long max(long a, long b) {
        if (a < b) {
            return b;
        }
        return a;
    }

    private static int max(int a, int b) {
        if (a < b) {
            return b;
        }
        return a;
    }

    private static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        public String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());

                }
            } catch (IOException e) {
                return null;
            }
            return tokenizer.nextToken();
        }

        public String nextLine() {
            String line = null;
            try {
                tokenizer = null;
                line = reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public boolean hasNext() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (Exception e) {
                return false;
            }
            return true;
        }
        
        public int[] nextIntArr(int n) {
        	int arr[] = new int[n];
    		for(int i = 0; i < n; i++) {
    			arr[i] = nextInt();
    		}
    		return arr;
        }
        
        public long[] nextLongArr(int n) {
        	long arr[] = new long[n];
    		for(int i = 0; i < n; i++) {
    			arr[i] = nextLong();
    		}
    		return arr;
        }
        
        public int[] nextIntArr1(int n) {
        	int arr[] = new int[n + 1];
    		for(int i = 1; i <= n; i++) {
    			arr[i] = nextInt();
    		}
    		return arr;
        }
        
        public long[] nextLongArr1(int n) {
        	long arr[] = new long[n + 1];
    		for(int i = 1; i <= n; i++) {
    			arr[i] = nextLong();
    		}
    		return arr;
        }
       
        public void close() {
        	try {
        		if(reader != null) {
        			reader.close();
        		}
        	}
        	catch(Exception e) {
        		
        	}
        	
        	
        }
    }

}