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>
#define int long long
#define ll long long
using ull = unsigned long long;
using namespace std;
#define dump(x)                             \
    if (dbg) {                              \
        cerr << #x << " = " << (x) << endl; \
    }
#define overload4(_1, _2, _3, _4, name, ...) name
#define FOR1(n) for (ll i = 0; i < (n); ++i)
#define FOR2(i, n) for (ll i = 0; i < (n); ++i)
#define FOR3(i, a, b) for (ll i = (a); i < (b); ++i)
#define FOR4(i, a, b, c) for (ll i = (a); i < (b); i += (c))
#define FOR(...) overload4(__VA_ARGS__, FOR4, FOR3, FOR2, FOR1)(__VA_ARGS__)
#define FORR(i, a, b) for (int i = (a); i <= (b); ++i)
#define bit(n, k) ((n >> k) & 1) /*nのk bit目*/
namespace mydef {
const int INF = 1ll << 60;
const int MOD = 1e9 + 7;
template <class T>
bool chmin(T& a, const T& b) {
    if (a > b) {
        a = b;
        return 1;
    } else
        return 0;
}
template <class T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return 1;
    } else
        return 0;
}
void Yes(bool flag = true) {
    if (flag)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
void No(bool flag = true) {
    Yes(!flag);
}
void YES(bool flag = true) {
    if (flag)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
void NO(bool flag = true) {
    YES(!flag);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T& val) {
    std::fill((T*)array, (T*)(array + N), val);
}
bool dbg = false;
}  // namespace mydef
using namespace mydef;
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(), (v).end()
#define SZ(x) ((int)(x).size())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vp vector<pair<int, int>>
#define vvp vector<vector<pair<int, int>>>
#define pi pair<int, int>
//#define P pair<int, int>
//#define V vector<int>
//#define S set<int>
#define asn ans

int N;
string S[100];
int C[100];
using P = pair<pair<string, int>, int>;

void solve() {
    int ans = INF;
    for (int i = 0; i < N; i++) {
        string s = S[i];
        reverse(all(s));
        if (S[i] == s)
            chmin(ans, C[i]);
    }
    auto compare = [](P x, P y) { return x.second > y.second; };
    priority_queue<P, vector<P>, decltype(compare)> que{compare};
    for (int i = 0; i < N; i++) {
        que.push(make_pair(make_pair(S[i], 1ll), C[i]));
    }
    int cnt = 0;
    while (!que.empty()) {
        cnt++;
        if (cnt >= 1e6) {
            cout << -1 << endl;
            exit(0);
        }
        auto p = que.top();
        que.pop();
        string s = p.first.first;
        int parity = p.first.second;
        int cost = p.second;
        if (cost >= ans)
            continue;
        for (int i = 0; i < N; i++) {
            auto nc = C[i];
            if (cost + nc >= ans)
                continue;
            auto ns = S[i];
            string s = p.first.first;
            if (parity == 1) {
                while (s.size() > 0 && ns.size() > 0) {
                    if (s.front() == ns.back()) {
                        s.erase(s.begin());
                        ns.pop_back();
                    } else
                        break;
                }
                if (s.size() + ns.size() <= 1) {
                    chmin(ans, cost + nc);
                } else if (s.size() == 0) {
                    que.push(make_pair(make_pair(ns, 0), cost + nc));
                } else if (ns.size() == 0) {
                    que.push(make_pair(make_pair(s, 1), cost + nc));
                }
            } else {
                while (s.size() > 0 && ns.size() > 0) {
                    if (ns.front() == s.back()) {
                        ns.erase(ns.begin());
                        s.pop_back();
                    } else
                        break;
                }
                if (s.size() + ns.size() <= 1) {
                    chmin(ans, cost + nc);
                } else if (s.size() == 0) {
                    que.push(make_pair(make_pair(ns, 1), cost + nc));
                } else if (ns.size() == 0) {
                    que.push(make_pair(make_pair(s, 0), cost + nc));
                }
            }
        }
    }
    if (ans == INF) {
        ans = -1;
    }
    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> N;
    for (int i = 0; i < N; i++) {
        cin >> S[i] >> C[i];
    }

    solve();
    return 0;
}