Codeforces #367（Div.2）C.Hard problem【dp】

C. Hard problem

time limit per test

 1 second

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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

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.

题目大意：

有n个字符串，每个字符串翻转都需要一定的花费，问最小花费使得全部字符串a【i】>=a【i-1】（当然是字典序上边的比较）

思路：

1、怀着侥幸心理敲了一发char【】【】.然后Wa 12，后来想测试一下是不是本地开不了那么大的数组到CF就行.......然后光荣的TLE............题干上给出的是所有字符串总长不超过100000，不代表某一个字符串就没有100000的时候。

2、然后开了Sring..............

设定dp【i】【2】-----------dp【i】【0】表示dp过程进行到第i个字符串并且当前字符串不翻转的最小花费。dp【i】【1】表示dp过程进行到第i个字符串并且当前字符串翻转的最小花费。

3、那么其状态转移不难写出：

dp【i】【0】=min（dp【i】【0】，dp【i-1】【0】）前提条件：a【i】>=a【i-1】；

dp【i】【0】=min（dp【i】【0】，dp【i-1】【1】）前提条件：a【i】>=reverse（a【i-1】）；

dp【i】【1】=min（dp【i】【1】，dp【i-1】【0】+val【i】）前提条件：reverse（a【i】）>=a【i-1】；

dp【i】【1】=min（dp【i】【1】，dp【i-1】【1】+val【i】）前提条件：reverse（a【i】）>=reverse（a【i-1】）；

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define ll __int64
#define INF 10000000000000000
string a[100005];
string b[100005];
ll dp[100005][2];
ll val[1000005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            dp[i][0]=INF;
            dp[i][1]=INF;
            scanf("%d",&val[i]);
        }
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            b[i]=a[i];
            reverse(b[i].begin(),b[i].end());
        }
        dp[1][0]=0;
        dp[1][1]=val[1];
        for(int i=2;i<=n;i++)
        {
            if(a[i]>=a[i-1])dp[i][0]=min(dp[i][0],dp[i-1][0]);
            if(a[i]>=b[i-1])dp[i][0]=min(dp[i][0],dp[i-1][1]);
            if(b[i]>=a[i-1])dp[i][1]=min(dp[i][1],dp[i-1][0]+val[i]);
            if(b[i]>=b[i-1])dp[i][1]=min(dp[i][1],dp[i-1][1]+val[i]);
        }
        ll ans=INF;
        ans=min(dp[n][0],dp[n][1]);
        if(ans>=INF)printf("-1\n");
        else printf("%I64d\n",ans);
    }
}