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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target( \
      "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
  using namespace std;
  const long long N = 15;
  const long long M = 2e6 + 5;
  const long long SZ = 550;
  const double eps = 1e-9;
  const long long mod = 1e9 + 7;
  const long long inf = 0x3f3f3f3f3f3f3f;
  long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  void wt(long long x) {
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (x >= 10) wt(x / 10), putchar(10);
    putchar(x % 10 + '0');
  }
  long long ksm(long long x, long long y = mod - 2, long long z = mod) {
    long long ret = 1;
    while (y) {
      if (y & 1) ret = (ret * x) % z;
      x = (x * x) % z;
      y >>= 1;
    }
    return ret;
  }
  long long inv[M], fac[M], ifc[M];
  void Init(long long n) {
    inv[1] = 1;
    for (long long i = 2; i <= n; i++)
      inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
    fac[0] = 1;
    for (long long i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
    ifc[0] = 1;
    for (long long i = 1; i <= n; i++) ifc[i] = 1ll * ifc[i - 1] * inv[i] % mod;
  }
  long long C(long long n, long long m) {
    if (m < 0 || m > n) return 0;
    return 1ll * fac[n] * ifc[m] % mod * ifc[n - m] % mod;
  }
  long long n, ans;
  long long a[N], g[N][N];
  long long f[1 << 14];
  vector<long long> v1, v2;
  signed main() {
    Init(M - 1);
    n = read();
    for (long long i = 0; i < n; i++) a[i] = read();
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < n; j++) g[i][j] = a[i] * inv[a[i] + a[j]] % mod;
    for (long long S = 1; S < (1 << n); S++) {
      f[S] = 1;
      for (long long T = (S - 1) & S; T; T = (T - 1) & S) {
        long long X = S - T;
        v1.clear();
        v2.clear();
        for (long long i = 0; i < n; i++) {
          if (T >> i & 1) v1.push_back(i);
          if (X >> i & 1) v2.push_back(i);
        }
        long long P = 1;
        for (long long i : v1)
          for (long long j : v2) P = P * g[i][j] % mod;
        f[S] = (f[S] - P * f[T] % mod) % mod;
      }
    }
    for (long long S = 1; S < (1 << n); S++) {
      v1.clear();
      v2.clear();
      for (long long i = 0; i < n; i++)
        if (S >> i & 1)
          v1.push_back(i);
        else
          v2.push_back(i);
      long long P = 1;
      for (long long i : v1)
        for (long long j : v2) P = P * g[i][j] % mod;
      ans = (ans + P * f[S] % mod * (long long)v1.size()) % mod;
    }
    cout << (ans + mod) % mod << "\n";
  }
  