Codeforces Round #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.

题解：给你n个串，要你求：要使这n个串的字典序从小到大，要消耗的费用最小是多少。改变字典序只能通过反串来改变。

dp[i][0]表示到第i个字符串，第i个字符串不翻转，能消耗的最小值。

dp[i][1]表示到第i个字符串，第i个字符串翻转，能消耗的最小值。

详细看代码。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<cmath>
#include<queue>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)  
#define rep(i,j,k) for (int i = j; i <= k; i++)  
#define per(i,j,k) for (int i = j; i >= k; i--)  
#define lson x << 1, l, mid  
#define rson x << 1 | 1, mid + 1, r  
const int lowbit(int x) { return x&-x; }  
const double eps = 1e-8;  
const int INF = 1e9+7; 
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;  
const ll mod = (1LL<<32);
const int N = 101010; 
template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }  
template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }
int read()
{
	int v = 0, f = 1;
	char c =getchar();
	while( c < 48 || 57 < c ){
		if(c=='-') f = -1;
		c = getchar();
	}
	while(48 <= c && c <= 57) 
		v = v*10+c-48, c = getchar();
	return v*f;
}
int n;
ll c[N];
string s1[N];
string s2[N];
ll dp[N][2];

int main() 
{
	#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    n=read();
    for (int i = 1; i <= n; i++) cin >> c[i];
    for (int i = 1; i <= n; i++) cin >> s1[i];
    for (int i = 1; i <= n; i++)
	{
        s2[i] = s1[i];
        reverse(s2[i].begin(),s2[i].end());
        
    }
    dp[1][0] = 0;
    dp[1][1] = c[1];
    for (int i = 2; i <= n; i++) 
	{
        dp[i][0] = inf;
        dp[i][1] = inf;
        if (s1[i-1] <= s1[i]) dp[i][0] = min(dp[i][0],dp[i-1][0]);
        if (s2[i-1] <= s1[i]) dp[i][0] = min(dp[i][0],dp[i-1][1]);
        if (s1[i-1] <= s2[i]) dp[i][1] = min(dp[i][1],dp[i-1][0]+c[i]);
        if (s2[i-1] <= s2[i]) dp[i][1] = min(dp[i][1],dp[i-1][1]+c[i]);
    }
    ll ans = min(dp[n][0],dp[n][1]);
    if (ans == inf) puts("-1");
    else printf("%I64d\n",ans);
}