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: Accepted)

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
#define owo(i,a, b) for(int i=(a);i<(b); ++i)
#define uwu(i,a, b) for(int i=(a)-1; i>=(b); --i)
#define senpai push_back
#define ttgl pair<int, int>
#define ayaya cout<<"ayaya~"<<endl
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<ttgl, null_type,less<ttgl>, rb_tree_tag,tree_order_statistics_node_update>*/
 
using ll = long long;
using ld = long double;
const ll MOD = 1000000007;
const ll root = 62;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
ll binpow(ll a,ll b){ll res=1;while(b){if(b&1)res=(res*a)%MOD;a=(a*a)%MOD;b>>=1;}return res;}
ll modInv(ll a){return binpow(a, MOD-2);}
const double PI = acos(-1);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 51;
const int mxC = 23;
//idea is we can define the center of the palindrome beforehand
//state is: (left/right), string id, last char we have processed
ll dist[2][mxN][mxC];
ll arr[mxN];
string at[mxN];
int n;
int main() {
    //freopen("file.in", "r", stdin);
    //freopen("file.out", "w", stdout);
    mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
    cin.tie(0)->sync_with_stdio(0);
    cin>>n;
    priority_queue<array<ll, 4>, vector<array<ll, 4>>, greater<array<ll, 4>>> pq;
    ll ans = INFLL;
    memset(dist, INFLL, sizeof(dist));
    owo(i, 0, n) {
        cin>>at[i]>>arr[i];
        owo(j, 0, at[i].size()) {
            bool ok = 1;
            int left = j-1;
            int right = j+1;
            while(ok&&left>=0&&right<=at[i].size()-1) {
                ok=at[i][left]==at[i][right];
                left--;
                right++;
            }
            if(ok) {
                if(left<0) {
                    dist[1][i][right] = arr[i];
                    pq.push({arr[i], 1, i, right});
                }
                if(right>=at[i].size()) {
                    dist[0][i][left+1] = arr[i];
                    pq.push({arr[i], 0, i, left+1});
                }
            }
        }
        int nxt = at[i].size();
        pq.push({arr[i], 0, i, nxt});
        pq.push({arr[i], 1, i, 0});
        dist[0][i][nxt] = arr[i];
        dist[1][i][0] = arr[i];
        owo(j, 1, at[i].size()) {
            bool ok = 1;
            int left = j-1;
            int right = j;
            while(ok&&left>=0&&right<=at[i].size()-1) {
                ok=at[i][left]==at[i][right];
                left--;
                right++;
            }
            if(ok) {
                if(left<0) {
                    dist[1][i][right] = arr[i];
                    pq.push({arr[i], 1, i, right});
                }
                if(right>=at[i].size()) {
                    dist[0][i][left+1] = arr[i];
                    pq.push({arr[i], 0, i, left+1});
                }
            }
        }
    }
    ll curr, dir, id, cnt;
    while(!pq.empty()) {
        auto [curr, dir, id, cnt] = pq.top();
        //cout<<curr<<" "<<dir<<" "<<id<<" "<<cnt<<"\n";
        pq.pop();
        if((dir==0&&cnt<=0)||(dir==1&&cnt>at[id].size()-1)) {
            ans= min(ans, curr);
            continue;
        }
        if(dir) {
            //add new string to left
            owo(i, 0, n) {
                int left = at[i].size()-1;
                int right = cnt;
                bool ok = true;
                while(ok&&left>=0&&right<=at[id].size()-1) {
                    ok=at[i][left]==at[id][right];
                    left--;
                    right++;
                }
                if(ok) {
                    if(left<0) {
                        if(curr+arr[i]<dist[1][id][right]) {
                            dist[1][id][right] = curr+arr[i];
                            pq.push({dist[1][id][right], 1, id, right});
                        }
                    }
                    if(right>=at[id].size()) {
                        if(curr+arr[i]<dist[0][i][left+1]) {
                            dist[0][i][left+1] = curr+arr[i];
                            pq.push({dist[0][i][left+1], 0, i, left+1});
                        }
                    }
                }
            }
        }else {
            //add new string to right
            owo(i, 0, n) {
                int left = cnt-1;
                int right = 0;
                bool ok = true;
                while(ok&&left>=0&&right<=at[i].size()-1) {
                    ok=at[id][left]==at[i][right];
                    left--;
                    right++;
                }
                if(ok) {
                    if(left<0) {
                        if(curr+arr[i]<dist[1][i][right]) {
                            dist[1][i][right] = curr+arr[i];
                            pq.push({dist[1][i][right], 1, i, right});
                        }
                    }
                    if(right>=at[i].size()) {
                        if(curr+arr[i]<dist[0][id][left+1]) {
                            dist[0][id][left+1] = curr+arr[i];
                            pq.push({dist[0][id][left+1], 0, id, left+1});
                        }
                    }
                }
            }
        }
    }
    cout<<(ans==INFLL ? -1 : ans)<<"\n";
    return 0;
}