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 c

i (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.

題解：當時寫這道題的時候對string類特別陌生，確切的說根本沒有用過，結果出現了各種問題，導致最後也沒有通過， 題意還是挺簡單的，就是給你n個字串，每個字串不超過10萬，你可以任意轉置字串，使得最後能夠使得這n個字 符串按字典序從小到大排列（位置不能改變），這裡的狀態轉移方程還是挺好推的。 dp[i][0]:代表第i個字串在不轉置的情況下的最優解 dp[i][1]:代表轉置當前字串的情況下的最優解 程式碼如下：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include <iostream>  
#include <stack>  
#include<algorithm>
using namespace std;
#define maxn 100005
#define inf  1e15
__int64 dp[maxn][2];
struct node
{
    string a;
	string b;
}stu[maxn];
__int64 ans[maxn];
char s[maxn];
char s1[maxn];
int  main()
{
	int i,j,n,m;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
           scanf("%I64d",&ans[i]);
    for(i=1;i<=n;i++)
	{
		scanf("%s",s);
		int len=strlen(s);
		s[len]='\0';
		int k=0;
		for(j=len-1;j>=0;j--)
		 s1[k++]=s[j];
		 s1[k]='\0';   
		 stu[i].a=s;
		 stu[i].b=s1;
	} 
	for(i=1;i<=n;i++)
	   dp[i][0]=dp[i][1]=inf;
	   dp[1][0]=0;dp[1][1]=ans[1];
	for(i=2;i<=n;i++)
	{
		if(stu[i].a>=stu[i-1].a)
		    dp[i][0]=dp[i-1][0];
		if(stu[i].b>=stu[i-1].a)
		  dp[i][1]=dp[i-1][0]+ans[i];
		if(stu[i].a>=stu[i-1].b)
		   dp[i][0]=min(dp[i][0],dp[i-1][1]);
		if(stu[i].b>=stu[i-1].b)
		   dp[i][1]=min(dp[i][1],dp[i-1][1]+ans[i]);
		if(dp[i][0]==inf && dp[i][1]==inf)
		  break;
	}
	if(i==n+1)
	   printf("%I64d\n",min(dp[n][0],dp[n][1]));
	else
	   printf("-1\n");
}