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>

const int MAX_N = 55;
const int MAX_S = 25;
const int64_t MIN = LONG_LONG_MIN/2;
int N;
std::string strs[MAX_N];
int64_t costs[MAX_N];
int64_t mem[MAX_N][MAX_S];
bool seen[MAX_N][MAX_S];
std::vector<int> cando[MAX_N][MAX_S];

int main() {
  for (int i = 0; i < MAX_N; ++i)
    for (int j = 0; j < MAX_S; ++j)
      mem[i][j] = LONG_LONG_MAX;
  std::cin >> N;
  for (int i = 0; i < N; ++i) {
    std::cin >> strs[i];
    std::cin >> costs[i];
  }
  for (int i = 0; i < N; ++i) {
    for (int si = 0; si < strs[i].size(); ++si) {
      for (int j = 0; j < N; ++j) {
        int sjsz = strs[j].size();
        bool doable = true;
        for (int offset = 0; offset < sjsz; ++offset) {
          if (si+offset >= strs[i].size()) break;
          if (strs[j][sjsz-offset-1] != strs[i][si+offset]) {
            doable = false;
            break;
          }
        }
        if (doable) {
          // printf("cand %d %d: %d\n", i, si, j);
          cando[i][si].push_back(j);
        }
      }
    }
  }

  std::set<std::tuple<int64_t, int, int>> q;
  for (int i = 0; i < N; ++i) {
    q.insert(std::make_tuple(costs[i], i, 0));
    mem[i][0] = costs[i];
  }

  int64_t best = LONG_LONG_MAX;
  while (q.size()) {
    auto vv = *q.begin();
    q.erase(q.begin());
    int64_t val = std::get<0>(vv);
    int s = std::get<2>(vv);
    int n = std::get<1>(vv);
    if (seen[n][s]) continue;
    seen[n][s] = true;

    if (s == strs[n].size() || s == strs[n].size() - 1) {
      mem[n][s] = val;
      best = std::min(best, val);
    }
    for (int cand : cando[n][s]) {
      int next_n, next_s;
      if (s + strs[cand].size() > strs[n].size()) {
        next_n = cand; next_s = strs[n].size() - s;
      } else {
        next_n = n; next_s = s + strs[cand].size();
      }
      if (!seen[next_n][next_s]) {
        int64_t v = val + costs[cand];
        if (v < mem[next_n][next_s]) {
          mem[next_n][next_s] = v;
          q.insert(std::make_tuple(v, next_n, next_s));
        }
      }
    }
  }

  printf("%lld\n", best < 0 || best == LONG_LONG_MAX ? -1ll : best);
}