Problem:
Score : 600 points

Problem Statement

We have N strings of lowercase English letters: S_1, S_2, \cdots, S_N.

Takahashi wants to make a string that is a palindrome by choosing one or more of these strings - the same string can be chosen more than once - and concatenating them in some order of his choice.

The cost of using the string S_i once is C_i, and the cost of using it multiple times is C_i multiplied by that number of times.

Find the minimum total cost needed to choose strings so that Takahashi can make a palindrome.

If there is no choice of strings in which he can make a palindrome, print -1.

Constraints

1 \leq N \leq 50

1 \leq |S_i| \leq 20

S_i consists of lowercase English letters.

1 \leq C_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
S_1 C_1
S_2 C_2
:
S_N C_N

Output

Print the minimum total cost needed to choose strings so that Takahashi can make a palindrome, or -1 if there is no such choice.

Sample Input 1

3
ba 3
abc 4
cbaa 5

Sample Output 1

7

We have ba, abc, and cbaa.

For example, we can use ba once and abc once for a cost of 7, then concatenate them in the order abc, ba to make a palindrome.
Also, we can use abc once and cbaa once for a cost of 9, then concatenate them in the order cbaa, abc to make a palindrome.

We cannot make a palindrome for a cost less than 7, so we should print 7.

Sample Input 2

2
abcab 5
cba 3

Sample Output 2

11

We can choose abcab once and cba twice, then concatenate them in the order abcab, cba, cba to make a palindrome for a cost of 11.

Sample Input 3

4
ab 5
cba 3
a 12
ab 10

Sample Output 3

8

We can choose a once, which is already a palindrome, but it is cheaper to concatenate ab and cba.

Sample Input 4

2
abc 1
ab 2

Sample Output 4

-1

We cannot make a palindrome, so we should print -1.

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1e9+5;
const ll oo=1e18+5;
const int N=2e5+5;
const int mod=1e9+7;

vector<vector<string>> s;
vector<int> cost;
ll dp[55][35][2];
int n;

bool isGud(const string& s,int l,int r){
	for(int i=l,j=r-1;i<j;++i,--j){
		if(s[i]!=s[j]) return 0;
	}
	return 1;
}

ll solve(int id,int suff,int dir){
	int len=(int)s[id][dir].size();
	if(isGud(s[id][dir],len-suff,len)) return 0;
	ll res=dp[id][suff][dir];
	if(res>=0) return res;
	res=INF;
	int rev=dir^1;
	for(int i=0;i<n;++i){
		int sizee=(int)s[i][rev].size();
		if(sizee<=suff){
			if(s[i][rev]==s[id][dir].substr(len-suff,sizee)){
				res=min(res,cost[i]+solve(id,suff-sizee,dir));
			}
		}
		else{
			if(s[id][dir].substr(len-suff)==s[i][rev].substr(0,suff)){
				res=min(res,cost[i]+solve(i,sizee-suff,rev));
			}
		}
	}
	return res;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	s.resize(n,vector<string>(2));
	cost.resize(n);
	for(int i=0;i<n;++i){
		cin>>s[i][0]>>cost[i];
		s[i][1]=s[i][0];
		reverse(all(s[i][1]));
	}
	ll ans=oo;	
	memset(dp,255,sizeof(dp));
	for(int i=0;i<n;++i){
		ll cur=cost[i]+solve(i,(int)s[i][0].size(),0);
		ans=min(ans,cur);
	}
	if(ans>=INF) ans=-1;
	cout<<ans;
	return 0;
}