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;
typedef pair <int , int> pii;
typedef long double ld;

const int N = 51 , mod = 1e9+7;

string s[N];
int c[N];
vector <int> g[N][N][2];
set < pair < pair <ll , int> , pii> > st;
ll ans = 1ll * mod * mod , dis[N][N][2];

void dijekstra(){
   while(st.size()){
      auto x = *st.begin();
      st.erase(st.begin());
      ll d = x.first.first;
      if(d >= ans)
         break;
      int v = x.first.second;
      int len = x.second.first;
      int pos = x.second.second;
      for(int u : g[v][len][pos]){
         if(abs(len - int(s[u].size())) <= 1){
            ans = min(ans , d + c[u]);
         }
         else if(len > s[u].length()){
            if(dis[v][len - s[u].length()][pos] > d + c[u]){
               st.erase({{dis[v][len - s[u].length()][pos] , v} , {len - s[u].length() , pos}});
               dis[v][len - s[u].length()][pos] = d + c[u];
               st.insert({{dis[v][len - s[u].length()][pos] , v} , {len - s[u].length() , pos}});
            }
         }
         else{
            if(dis[u][s[u].length() - len][1 - pos] > d + c[u]){
               st.erase({{dis[u][s[u].length() - len][1 - pos] , v} , {s[u].length() - len , 1 - pos}});
               dis[u][s[u].length() - len][1 - pos] = d + c[u];
               st.insert({{dis[u][s[u].length() - len][1 - pos] , v} , {s[u].length() - len , 1 - pos}});
            }
         }
      }
   }
}

int32_t main(){
   ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
   int n;
   cin >> n;
   for(int i = 0 ; i < n ; i++){
      cin >> s[i] >> c[i];
      if(s[i].length() == 1)
         ans = min(ans , 1ll * c[i]);
   }
   for(int i = 0 ; i < n ; i++){
      for(int j = 1 ; j <= s[i].length() ; j++){
         for(int k = 0 ; k < n ; k++){
            bool bad = false;
            for(int p = 0 ; p < min(j , int(s[k].length())) ; p++){
               if(s[i][s[i].length() - j + p] != s[k][s[k].length() - 1 - p])
                  bad = true;
            }
            if(!bad)
               g[i][j][0].push_back(k);

            bad = false;
            for(int p = 0 ; p < min(j , int(s[k].length())) ; p++){
               if(s[i][j - 1 - p] != s[k][p])
                  bad = true;
            }
            if(!bad)
               g[i][j][1].push_back(k);
         }
      }
   }
   for(int i = 0 ; i < N ; i++)
      for(int j = 0 ; j < N ; j++)
         for(int k = 0 ; k < 2 ; k++)
            dis[i][j][k] = 1ll * mod * mod;
   for(int i = 0 ; i < n ; i++){
      st.insert({{c[i] , i} , {s[i].length() , 0}});
      dis[i][s[i].length()][0] = c[i];
   }
   dijekstra();
   if(ans == 1ll * mod * mod)
      ans = -1;
   cout << ans;
   return 0;
}