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;

#define NDEBUG
#include <cassert>


typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef pair<ll, ll> llll;
typedef pair<double, double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<llll> vllll;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a) int((a).size())
#define pb push_back
#define eb emplace_back
#define FOR(var, from, to) for (int var = (from); var <= (to); ++var)
#define rep(var, n) for (int var = 0; var < (n); ++var)
#define rep1(var, n) for (int var = 1; var <= (n); ++var)
#define repC2(vari, varj, n)                 \
    for (int vari = 0; vari < (n)-1; ++vari) \
        for (int varj = vari + 1; varj < (n); ++varj)
#define repC3(vari, varj, vark, n)                      \
    for (int vari = 0; vari < (n)-2; ++vari)            \
        for (int varj = vari + 1; varj < (n)-1; ++varj) \
            for (int vark = varj + 1; vark < (n); ++vark)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define tr(i, c) for (auto i = (c).begin(); i != (c).end(); ++i)
#define found(s, e) ((s).find(e) != (s).end())
#define mset(arr, val) memset(arr, val, sizeof(arr))
#define mid(x, y) ((x) + ((y) - (x)) / 2)
#define IN(x, a, b) ((a) <= (x) && (x) <= (b))
#define cons make_pair
#define clamp(v, lo, hi) min(max(v, lo), hi)
#define ABS(x) max((x), -(x))
#define PQ(T) priority_queue<T, vector<T>, greater<T>>

template <typename T1, typename T2>
inline void amin(T1& a, T2 const& b) {
    if (a > b) a = b;
}
template <typename T1, typename T2>
inline void amax(T1& a, T2 const& b) {
    if (a < b) a = b;
}
template <typename X, typename T>
auto vectors(X x, T a) {
    return vector<T>(x, a);
}
template <typename X, typename Y, typename Z, typename... Zs>
auto vectors(X x, Y y, Z z, Zs... zs) {
    auto cont = vectors(y, z, zs...);
    return vector<decltype(cont)>(x, cont);
}

inline ll square(ll x) { return x * x; }
inline ll gcd(ll a, ll b) {
    while (a) swap(a, b %= a);
    return b;
}
inline ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template <typename T>
inline T mod(T a, T b) {
    return ((a % b) + b) % b;
}

template <typename T>
int find_left(vector<T>& v, T elem) {
    return (int)(upper_bound(v.begin(), v.end(), elem) - v.begin()) - 1;
}
template <typename T>
int find_right(vector<T>& v, T elem) {
    return (int)(lower_bound(v.begin(), v.end(), elem) - v.begin());
}

const ll MOD = 1000000007LL;

inline ll ADD(ll x, ll y) { return (x + y) % MOD; }
inline ll SUB(ll x, ll y) { return (x - y + MOD) % MOD; }
inline ll MUL(ll x, ll y) { return x * y % MOD; }
inline ll POW(ll x, ll e) {
    ll v = 1;
    for (; e; x = MUL(x, x), e >>= 1)
        if (e & 1) v = MUL(v, x);
    return v;
}
inline ll INV(ll y) { /*assert(y%MOD!=0);*/
    return POW(y, MOD - 2);
}
inline ll DIV(ll x, ll y) { return MUL(x, INV(y)); }

#define INTSPACE 12
char _buf[INTSPACE * 1000000 + 3];

int loadint() {
    if (fgets(_buf, INTSPACE + 3, stdin) == NULL) return 0;
    return atoi(_buf);
}

int loadvec(vector<int>& v, int N = -1) {
    if (N == 0) {
        v.clear();
        return 0;
    }
    if (N == -1) {
        N = loadint();
        if (N == 0) return 0;
    }
    int bufsize = INTSPACE * N + 3;
    if (fgets(_buf, bufsize, stdin) == NULL) return 0;
    v.resize(N);

    int i = 0;
    bool last = false;
    for (char* p = &_buf[0];;) {
        char* q = p;
        while (*q > ' ') ++q;
        if (*q == 0x0D || *q == 0x0A) last = true;
        *q = 0;
        v[i++] = atoi(p);
        if (last || i == N) break;
        p = q + 1;
    }
    return i;
}
void read_cr() { fgets(_buf, 256, stdin); }

void horizontal(vector<int>& v) {
    int L = v.size();
    for (int i = 0; i < L; ++i) {
        printf("%d%c", v[i], (i == L - 1) ? '\n' : ' ');
    }
}
void horizontall(vector<long long>& v) {
    int L = v.size();
    for (int i = 0; i < L; ++i) {
        printf("%lld%c", v[i], (i == L - 1) ? '\n' : ' ');
    }
}

string rev(string s) { return string(s.rbegin(), s.rend()); }

using P = tuple<ll, string, bool>;
using Q = pair<string, bool>;

inline bool palin(string& s) {
    int L = s.size();
    if (L < 2) return true;
    for (int i = 0, j = L - 1; i < j; ++i, --j)
        if (s[i] != s[j]) return false;
    return true;
}
inline int common(string& s, string& t) {
    int L = min(s.size(), t.size()), c = 0;
    rep(i, L) {
        if (s[i] != t[i]) return i;
    }
    return L;
}

ll solve(int N, vs& s, vi& c) {
    vs r(N);
    ll best = LLONG_MAX;
    rep(i, N) {
        r[i] = rev(s[i]);
    }

    PQ(P) pq;
    map<Q, ll> distance;
    rep(i, N) { pq.emplace((ll)c[i], s[i], false); }

    while (!pq.empty()) {
        auto [ct, st, rt] = pq.top();
        pq.pop();
        if (ct > best) continue;

        if (palin(st)) {
            amin(best, ct);
            continue;
        }

        Q q(st, rt);
        if (found(distance, q)) {
            if (distance[q] <= ct) continue;
        }
        distance[q] = ct;

        rep(i, N) {
            string zt = rt ? s[i] : r[i];
            ll newcost = ct + c[i];
            int com = common(st, zt);
            if (com != min(st.size(), zt.size())) continue;
            if (st.size() >= zt.size()) {
                pq.emplace(newcost, st.substr(com), rt);
            } else {
                pq.emplace(newcost, rev(zt.substr(com)), !rt);
            }
        }
    }

    return (best == LLONG_MAX) ? -1 : best;
}

int main() {
    int N;
    cin >> N;
    vs s(N);
    vi c(N);
    rep(i, N) cin >> s[i] >> c[i];
    cout << solve(N, s, c) << endl;
    return 0;
}