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;
typedef long long ll;

#define INF 1000000000000000LL

bool kaibun(string s) {
    int len = s.length();
    for (int i = 0; i < len; ++i) {
        if (s[i] != s[len-1-i]) {
            return false;
        }
    }
    return true;
}

int main() {
    int N; cin >> N;
    map<string, ll> SC;
    for (int i = 0; i < N; ++i) {
        string s; ll c; cin >> s >> c;
        if (SC[s] != 0) SC[s] = min(SC[s], c);
        else SC[s] = c;
    }
    bool updated;
    do {
        updated = false;
        map<string, ll> nextSC = SC;
        for (auto i = SC.begin(); i != SC.end(); ++i) {
            for (auto j = SC.begin(); j != SC.end(); ++j) {
                string is, js;
                ll ic, jc;
                is = i->first; js = j->first;
                ic = i->second; jc = j->second;
                int isl = is.length();
                int jsl = js.length();
                string ream = "";
                if (isl > jsl) {
                    ream = is.substr(jsl);
                } else if (isl < jsl) {
                    ream = js.substr(0, jsl-isl);
                }
                bool ok = true;
                for (int idx = 0; idx < min(isl, jsl); ++idx) {
                    if (is[idx] != js[jsl-1-idx]) {
                        ok = false;
                        break;
                    }
                }
                if (ok) {
                    if (nextSC[ream] == 0) {
                        updated = true;
                        nextSC[ream] = ic + jc;
                    } else if (nextSC[ream] > ic + jc) {
                        updated = true;
                        nextSC[ream] = ic + jc;
                    }
                }
            }
        }
        SC = nextSC;
    } while(updated);
    ll ans = INF;
    for (auto itr = SC.begin(); itr != SC.end(); ++itr) {
        string s = itr->first;
        ll c = itr->second;
        if (kaibun(s)) ans = min(ans, c);
    }
    if (ans == INF) {
        cout << -1 << endl;
    } else {
        cout << ans << endl;
    }
}