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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define ff first
#define ss second
#define PI acos(-1)
#define pb push_back
#define int long long
#define ld long double
#define sp fixed<<setprecision
#define bp __builtin_popcountll
#define all(x) x.begin(),x.end()
#define pii pair<long long,long long>
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

const int M = (1e9)+7;
const int N = 1e6+5;

bool ok(string x)
{
    int n = x.length();
    for(int i=0;i<n/2;i++)if(x[i] != x[n-1-i])return false;
    return true;
}

signed main()
{
    FAST
    int tc=1;
    //cin>>tc;
    for(int ti=1;ti<=tc;ti++)
    {
        int n,ans = 1e15;
        cin>>n;
        string s[n];
        int c[n];
        for(int i=0;i<n;i++)cin>>s[i]>>c[i];
        for(int i=0;i<n;i++)if(ok(s[i]))ans = min(ans,c[i]);
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)
        {
            string chk = s[i]+s[j];
            if(ok(chk))ans = min(ans,c[i]+c[j]);
        }
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)for(int k=0;k<n;k++)
        {
            string chk = s[i]+s[j]+s[k];
            if(ok(chk))ans = min(ans,c[i]+c[j]+c[k]);
        }
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)for(int k=0;k<n;k++)for(int l=0;l<n;l++)
        {
            string chk = s[i]+s[j]+s[k]+s[l];
            if(ok(chk))ans = min(ans,c[i]+c[j]+c[k]+c[l]);
        }
        if(ans == 1e15)ans = -1;
        cout<<ans<<endl;
    }
    return 0;
}