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 <algorithm>
#include <climits>
#include <iostream>
#include <queue>
#include <set>
#include <unordered_map>
#include <vector>

using namespace std;
typedef long long ll;

bool is_palindrome(string &s) {
    int n = s.size();
    for (int i = 0; i < n / 2; ++i)
        if (s[i] != s[n - i - 1])
            return false;
    return true;
}

int main() {
    int n;
    cin >> n;
    priority_queue<tuple<ll, string, bool>, vector<tuple<ll, string, bool>>,
        greater<>>
        pq;
    unordered_map<string, ll> dp, ds;
    ll ans = LLONG_MAX;
    for (int i = 0; i < n; ++i) {
        string s;
        ll c;
        cin >> s >> c;
        if (is_palindrome(s))
            ans = min(ans, c);
        pq.push({c, s, true});
        if (!dp.count(s))
            dp[s] = c;
        else
            dp[s] = min(dp[s], c);
        reverse(s.begin(), s.end());
        if (!ds.count(s))
            ds[s] = c;
        else
            ds[s] = min(ds[s], c);
    }
    set<pair<string, bool>> vis;
    auto merge = [&](string prefix, string suffix, ll nc) {
        int lp = prefix.size();
        int ls = suffix.size();
        int l = min(lp, ls);
        string pp = prefix.substr(0, l);
        string ps = suffix.substr(0, l);
        if (pp != ps)
            return;
        if (lp > l) {
            string rp = prefix.substr(l, lp - l);
            if (is_palindrome(rp))
                ans = min(ans, nc);
            else if (!dp.count(rp) || dp[rp] > nc) {
                dp[rp] = nc;
                pq.push({nc, rp, true});
            }
        } else {
            string rs = suffix.substr(l, ls - l);
            if (is_palindrome(rs))
                ans = min(ans, nc);
            else if (!ds.count(rs) || ds[rs] > nc) {
                ds[rs] = nc;
                pq.push({nc, rs, false});
            }
        }
    };
    while (!pq.empty()) {
        auto[cost, s, is_prefix] = pq.top();
        pq.pop();
        if (cost >= ans)
            break;
        if (vis.count({s, is_prefix}))
            continue;
        vis.insert({s, is_prefix});
        if (is_prefix) {
            for (auto &[suffix, dist] : ds)
                merge(s, suffix, cost + dist);
        } else {
            for (auto &[prefix, dist] : dp)
                merge(prefix, s, cost + dist);
        }
    }
    cout << (ans == LLONG_MAX ? -1 : ans) << endl;
}