codeforces 706C C. Hard problem(dp)

题目链接：

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 exceed100 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

input

3
1 3 1
aa
ba
ac

output

input

2
5 5
bbb
aaa

output

-1

input

2
3 3
aaa
aa

output

-1

题意：

问把这些字符串按字典序排好，如果第i个需要倒转，花费为a[i]，现在求最小花费;

思路：

dp[i][j]表示第i个串的状态为j时的最小花费,j==0表示不倒转,j==1表示倒转;转移方程看代码;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <bits/stdc++.h>
using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=998244353;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+10;
const int maxn=1e3+10;
const  double eps=1e-4;

int n;
LL a[N],dp[N][2];
string s[N];

int main()
{
        read(n);
        For(i,1,n)read(a[i]);
        For(i,0,n)dp[i][0]=dp[i][1]=inf;
        For(i,1,n)
        {
            cin>>s[i];
            if(i==1){dp[i][0]=0,dp[i][1]=a[1];continue;}
            if(s[i]>=s[i-1])dp[i][0]=min(dp[i][0],dp[i-1][0]);
            string temp="";
            int len=s[i-1].length();
            for(int j=len-1;j>=0;j--)
            {
                temp=temp+s[i-1][j];
            }
            if(s[i]>=temp)dp[i][0]=min(dp[i][0],dp[i-1][1]);
            string t="";
            len=s[i].length();
            for(int j=len-1;j>=0;j--)t=t+s[i][j];
            if(t>=s[i-1])dp[i][1]=min(dp[i][1],dp[i-1][0]+a[i]);
            if(t>=temp)dp[i][1]=min(dp[i][1],dp[i-1][1]+a[i]);
        }
        LL ans=min(dp[n][0],dp[n][1]);
        if(ans==inf)cout<<"-1";
        else cout<<ans;
        return 0;
}