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;

int n;
string s[54];
long long c[54];
long long dist[54][2][24];
priority_queue<pair<long long, pair<int, pair<int, int>>>> pq;
vector<long long> ans;

void solve(){
    cin>>n;
    memset(dist, -1, sizeof(dist));
    for(int i = 0; i < n; i++){
        cin>>s[i]>>c[i];
        dist[i][0][s[i].length()] = c[i];
        pq.push({-c[i], {i, {0, s[i].length()}}});
    }
    while(!pq.empty()){
        pair<long long, pair<int, pair<int, int>>> now = pq.top();
        pq.pop();
        long long cost = -now.first;
        int idx = now.second.first;
        int dir = now.second.second.first;
        int rem = now.second.second.second;
        if(cost > dist[idx][dir][rem])  continue;
        if(dir == 0){
            for(int nextIdx = 0; nextIdx < n; nextIdx++){
                const int L = s[idx].length();
                const int R = s[nextIdx].length();
                bool can = true;
                int lRem = rem, rRem = R;
                while(lRem && rRem){
                    if(s[idx][L-lRem] != s[nextIdx][rRem-1])  can = false;
                    lRem--;
                    rRem--;
                }
                if(!can)    continue;
                long long nextCost = cost + c[nextIdx];
                if(lRem + rRem <= 1){
                    ans.push_back(nextCost);
                    continue;
                }
                if(lRem && (dist[idx][0][lRem] > nextCost || dist[idx][0][lRem] == -1)){
                    dist[idx][0][lRem] = nextCost;
                    pq.push({-nextCost, {idx, {0, lRem}}});
                }
                if(rRem && (dist[nextIdx][1][rRem] > nextCost || dist[nextIdx][1][rRem] == -1)){
                    dist[nextIdx][1][rRem] = nextCost;
                    pq.push({-nextCost, {nextIdx, {1, rRem}}});
                }
            }
        }
        else {
            for(int nextIdx = 0; nextIdx < n; nextIdx++){
                const int L = s[nextIdx].length();
                const int R = s[idx].length();
                bool can = true;
                int lRem = L, rRem = rem;
                while(lRem && rRem){
                    if(s[nextIdx][L-lRem] != s[idx][rRem-1])  can = false;
                    lRem--;
                    rRem--;
                }
                if(!can)    continue;
                long long nextCost = cost + c[nextIdx];
                if(lRem + rRem <= 1){
                    ans.push_back(nextCost);
                    continue;
                }
                if(lRem && (dist[nextIdx][0][lRem] > nextCost || dist[nextIdx][0][lRem] == -1)){
                    dist[nextIdx][0][lRem] = nextCost;
                    pq.push({-nextCost, {nextIdx, {0, lRem}}});
                }
                if(rRem && (dist[idx][1][rRem] > nextCost || dist[idx][1][rRem] == -1)){
                    dist[idx][1][rRem] = nextCost;
                    pq.push({-nextCost, {idx, {1, rRem}}});
                }
            }
        }
    }
    if(ans.size() == 0) cout<<-1<<endl;
    else
    {
        cout<<*min_element(ans.begin(), ans.end());
    }
    
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int TC;
    //cin>>TC;
    TC=1;
    while(TC--){
        solve();
    }
}