description: <image>
  
  William is not only interested in trading but also in betting on sports matches. n teams participate in each match. Each team is characterized by strength a_i. Each two teams i < j play with each other exactly once. Team i wins with probability (a_i)/(a_i + a_j) and team j wins with probability (a_j)/(a_i + a_j).
  
  The team is called a winner if it directly or indirectly defeated all other teams. Team a defeated (directly or indirectly) team b if there is a sequence of teams c_1, c_2, ... c_k such that c_1 = a, c_k = b and team c_i defeated team c_{i + 1} for all i from 1 to k - 1. Note that it is possible that team a defeated team b and in the same time team b defeated team a.
  
  William wants you to find the expected value of the number of winners.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 14), which is the total number of teams participating in a match.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — the strengths of teams participating in a match.
  
  Output
  
  Output a single integer — the expected value of the number of winners of the tournament modulo 10^9 + 7.
  
  Formally, let M = 10^9+7. It can be demonstrated that the answer can be presented as a irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output a single integer equal to p ⋅ q^{-1} mod M. In other words, output an integer x such that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.
  
  Examples
  
  Input
  
  
  2
  1 2
  
  
  Output
  
  
  1
  
  
  Input
  
  
  5
  1 5 2 11 14
  
  
  Output
  
  
  642377629
  
  Note
  
  To better understand in which situation several winners are possible let's examine the second test:
  
  One possible result of the tournament is as follows (a → b means that a defeated b):
  
    * 1 → 2 
    * 2 → 3 
    * 3 → 1 
    * 1 → 4 
    * 1 → 5 
    * 2 → 4 
    * 2 → 5 
    * 3 → 4 
    * 3 → 5 
    * 4 → 5 
  
  
  
  Or more clearly in the picture:
  
  <image>
  
  In this case every team from the set \{ 1, 2, 3 \} directly or indirectly defeated everyone. I.e.:
  
    * 1st defeated everyone because they can get to everyone else in the following way 1 → 2, 1 → 2 → 3, 1 → 4, 1 → 5. 
    * 2nd defeated everyone because they can get to everyone else in the following way 2 → 3, 2 → 3 → 1, 2 → 4, 2 → 5. 
    * 3rd defeated everyone because they can get to everyone else in the following way 3 → 1, 3 → 1 → 2, 3 → 4, 3 → 5. 
  
  
  
  Therefore the total number of winners is 3.
solution: #include <bits/stdc++.h>
  using namespace std;
  constexpr int MAXN = 14;
  int arr[MAXN];
  int masks[1 << MAXN];
  int add_mod(int a, int b) { return (a + b) % 1000000007; }
  int mult_mod(int a, int b) { return (int)((((long long)a) * b) % 1000000007); }
  int neg_mod(int a) { return (1000000007 - a) % 1000000007; }
  long long powmod(int base, int e) {
    if (e == 0) return 1;
    long long tmp = powmod(base, e / 2);
    tmp *= tmp;
    tmp %= 1000000007;
    if (e & 1) {
      tmp *= base;
      tmp %= 1000000007;
    }
    return tmp;
  }
  int inv(int a) { return (int)powmod(a, 1000000007 - 2); }
  int divi(int a, int b) { return mult_mod(a, inv(b)); }
  int prob[MAXN][MAXN];
  int noneness_cache[MAXN][1 << MAXN];
  int32_t main() {
    ios_base::sync_with_stdio(false);
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        prob[i][j] = divi(arr[i], add_mod(arr[i], arr[j]));
      }
    }
    for (int i = 0; i < n; ++i) {
      for (unsigned bm = 0; bm < (1u << n); ++bm) {
        int noneness = 1;
        for (int k = 0; k < n; ++k) {
          if ((1u << k) & bm) {
            noneness = mult_mod(noneness, prob[i][k]);
          }
        }
        noneness_cache[i][bm] = add_mod(1, neg_mod(noneness));
      }
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      for (unsigned bm = 0; bm != (1u << n); ++bm) {
        if ((bm & (1u << i)) == 0) {
          masks[bm] = 0;
        } else if (bm == (1u << i)) {
          masks[bm] = 1;
        } else {
          int total = 0;
          auto bm_popcnt = __builtin_popcount(bm);
          for (unsigned b2 = (bm - 1) & bm; b2 != 0; b2 = (b2 - 1) & bm) {
            int eff = masks[b2];
            unsigned diff = bm ^ b2;
            for (int j = 0; j < n; ++j) {
              if ((1 << j) & diff) {
                eff = mult_mod(eff, noneness_cache[j][b2]);
              }
            }
            auto b2_popcnt = __builtin_popcount(b2);
            if ((bm_popcnt - b2_popcnt) & 1) {
              total = add_mod(total, eff);
            } else {
              total = add_mod(total, neg_mod(eff));
            }
          }
          masks[bm] = total;
        }
      }
      ans = add_mod(ans, masks[(1u << n) - 1u]);
      for (unsigned bm = 0; bm != (1u << n); ++bm) {
        masks[bm] = 0;
      }
    }
    cout << ans << '\n';
  }
  