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: Runtime Error)

#include <bits/stdc++.h>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<P, ll> T;
typedef pair<long double , ll> Ps;
typedef pair<ll, bool> Pb;

const ll INF = 1e18;
const ll fact_table = 3200008;
long double Pi = 3.1415926535897932384626;

priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll, vector<ll>, greater<ll> > pqls;
priority_queue <P, vector<P>, greater<P> > pqps;
//small priority queue
//top pop

ll dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
ll dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
//↓,→,↑,←


#define endl "\n"
#ifdef ENJAPMA
#undef endl
#endif

#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)();
#define pc(x) cout << x << ",";
#define rep(i, n) for (ll i = 0; i < (n); i ++)

// const ll mod = 998244353ll;
const ll mod = 1000000007ll;

ll mypow(ll a, ll b, ll mod) {ll x = 1; while (b) {while (!(b & 1)) {(a *= a) %= mod; b >>= 1;}(x *= a) %= mod; b--;} return x;}
void YES(bool cond) { if (cond) { p("YES");} else { p("NO");} return;}
void Yes(bool cond) { if (cond) { p("Yes");} else { p("No");} return;}
void line() { p("--------------------"); return;}

/*
ll fact[fact_table + 5], rfact[fact_table + 5];

void c3_init() {
    fact[0] = rfact[0] = 1;
    for (ll i = 1; i <= fact_table; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
    }
    rfact[fact_table] = mypow(fact[fact_table], mod - 2, mod);
    for (ll i = fact_table; i >= 1; i--) {
        rfact[i - 1] = rfact[i] * i;
        rfact[i - 1] %= mod;
    }
    return;
}
ll c3(ll n, ll r) {
    return (((fact[n] * rfact[r]) % mod ) * rfact[n - r]) % mod;
}
*/
struct Timer {
    int64_t start;
    const int64_t CYCLES_PER_SEC = 2800000000;
    Timer() { reset(); }
    void reset() { start = getCycle(); }
    inline double get() { return (double)(getCycle() - start) / CYCLES_PER_SEC; }
    inline int64_t getCycle() {
        uint32_t low, high;
        __asm__ volatile ("rdtsc" : "=a" (low), "=d" (high));
        return ((int64_t)low) | ((int64_t)high << 32);
    }
};


bool multicase = false;

ll n, m, k, w, num, a, b, c, d, e, h, q, ans, sum, l;

typedef vector<ll> vec;
typedef vector<vector<ll>> mat;

ll dp[50][50][3];
string s[500];
ll cost[50];

ll calc(ll id, ll idx, int dir) {
    if (dp[id][idx][dir] != -1) return dp[id][idx][dir];
    // 残ってる文字列が怪文である場合の処理をしないといけません

    if (dir == 0) {
        if (idx == s[id].size()) return 0ll;
        string c = s[id].substr(idx, s[id].size() - idx);
        string d = c;
        reverse(d.begin(), d.end());
        if (c == d) return 0ll;
    }

    if (dir == 1) {
        if (idx == -1) return 0ll;
        string c = s[id].substr(0, idx + 1);
        string d = c;
        reverse(d.begin(), d.end());
        if (c == d) return 0ll;
    }

    ll res = INF;
    if (dir == 0) {
        string a = s[id].substr(idx, s[id].size() - idx);
        for(int i=0;i<n;i++){
            string b = s[i];
            if (a.size() > b.size()) {
                reverse(b.begin(), b.end());
                if (a.substr(0, b.size()) == b) {
                    dp[id][idx][dir] = calc(id, idx + b.size(), 0) + cost[i];
                    res = min(res, dp[id][idx][dir]);
                }
            } else {
                reverse(a.begin(), a.end());
                if (b.substr(b.size() - a.size(), a.size()) == a){
                    dp[id][idx][dir] = calc(i, b.size() - a.size() - 1, 1) + cost[i];
                    res = min(res, dp[id][idx][dir]);
                }
                reverse(a.begin(), a.end());
            }
        }
    } else {
        string a = s[id].substr(0, idx + 1);
        for(int i=0;i<n;i++){
            string b = s[i];
            if (a.size() > b.size()) {
                reverse(b.begin(), b.end());
                if (a.substr(a.size() - b.size(), b.size()) == b) {
                    dp[id][idx][dir] = calc(id, idx - b.size(), 1) + cost[i];
                    res = min(res, dp[id][idx][dir]);
                }
            } else {
                reverse(a.begin(), a.end());
                if (b.substr(0, a.size()) == a) {
                    dp[id][idx][dir] = calc(i, a.size(), 0) + cost[i]; 
                    res = min(res, dp[id][idx][dir]);
                }
                reverse(a.begin(), a.end());
            }
        }
    }
    return res;
}

void solve(){
    cin >> n;
    for(int i=0;i<n;i++){
        cin >> s[i] >> cost[i];
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<25;j++){
            dp[i][j][0] = dp[i][j][1] = -1;
        }
    }
    ll ans = INF;
    for(int i=0;i<n;i++){
        ans = min(ans, calc(i, 0, 0) + cost[i]);
    }
    if (ans == INF) {
        p(-1);
    }else{
        p(ans);
    }
    return;
}

int main() {
    // init();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll q, testcase = 1;
    if (multicase) {
        cin >> q;
    } else {
        q = 1;
    }
    while (q--) {
        // pu("Case ");pu("#");pu(testcase);pu(": ");
        solve();
        testcase++;
    }
    // solve();
    return 0;
}