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;
  template <typename T1, typename T2>
  inline void chkmin(T1 &a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  inline void chkmax(T1 &a, T2 b) {
    if (a < b) a = b;
  }
  using ll = long long;
  using ld = long double;
  const string FILENAME = "input";
  const int Mod = 1000000007;
  int sum(int a, int b) { return (a + b >= Mod ? a + b - Mod : a + b); }
  int mul(int a, int b) { return ((ll)a * b) % Mod; }
  int powm(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) {
        res = mul(res, a);
      }
      a = mul(a, a);
      b >>= 1;
    }
    return res;
  }
  int inv(int a) { return powm(a, Mod - 2); }
  const int MAXN = 15;
  int n;
  int a[MAXN];
  int p[MAXN][MAXN];
  int dp[1 << MAXN][MAXN][MAXN];
  int who[MAXN];
  unordered_map<int, int> val[1 << MAXN];
  int whos[1 << MAXN];
  int kek[1 << MAXN];
  int g[6000228];
  int pref[1 << MAXN];
  int pos[1 << MAXN];
  void solve() {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        p[i][j] = mul(a[i], powm(a[i] + a[j], Mod - 2));
      }
      p[i][i] = 1;
    }
    int uk = 0;
    for (int mask = 0; mask < (1 << n); mask++) {
      vector<int> gg;
      for (int t = 0; t < n; t++) {
        if (mask & (1 << t)) {
          gg.push_back(t);
          who[t] = (int)(gg).size() - 1;
        }
      }
      pref[mask] = uk;
      for (int mask1 = 0; mask1 < (1 << (int)(gg).size()); mask1++) {
        int sub = 0;
        for (int k = 0; k < (int)(gg).size(); k++) {
          if (mask1 & (1 << k)) {
            sub |= (1 << gg[k]);
          }
        }
        whos[mask1] = sub;
        int nsub = mask ^ sub;
        int res = 1;
        for (int k = 0; k < n; k++) {
          if (sub & (1 << k)) {
            for (int k1 = 0; k1 < n; k1++) {
              if (nsub & (1 << k1)) {
                res = mul(res, p[k1][k]);
              }
            }
          }
        }
        g[uk] = res;
        pos[sub] = uk - pref[mask];
        uk++;
      }
      for (auto i : gg) {
        if ((int)(gg).size() == 1) {
          for (int j = 0; j <= n; j++) {
            dp[mask][i][j] = 1;
          }
        } else {
          for (int mask1 = 0; mask1 < (1 << (int)(gg).size()); mask1++) {
            int sub = whos[mask1];
            if (sub & (1 << i)) {
              continue;
            }
            int nsub = mask ^ (1 << i) ^ sub;
            int npos = 0;
            int tp = 1;
            for (int k = 0; k < (int)(gg).size(); k++) {
              if (sub & (1 << gg[k])) {
                npos += tp;
              }
              if ((sub | nsub) & (1 << gg[k])) {
                tp <<= 1;
              }
            }
            kek[mask1] = g[pref[sub | nsub] + npos];
          }
          for (int j = n - 1; j >= 0; j--) {
            if (i != j && (mask & (1 << j))) {
              for (int mask1 = 0; mask1 < (1 << (int)(gg).size()); mask1++) {
                int sub = whos[mask1];
                if (sub & (1 << i)) {
                  continue;
                }
                if (!(sub & (1 << j))) {
                  continue;
                }
                int nsub = mask ^ (1 << i) ^ sub;
                dp[mask][i][j] =
                    sum(dp[mask][i][j],
                        mul(kek[mask1],
                            mul(dp[sub][j][0], dp[nsub | (1 << i)][i][j + 1])));
              }
              dp[mask][i][j] = mul(dp[mask][i][j], p[i][j]);
            }
            dp[mask][i][j] =
                sum(dp[mask][i][j],
                    mul(dp[mask][i][j + 1],
                        (i == j || !(mask & (1 << j)) ? 1 : p[j][i])));
          }
        }
      }
    }
    int ans = 0;
    for (int t = 0; t < n; t++) {
      ans = sum(ans, dp[(1 << n) - 1][t][0]);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  