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>
  template <class T>
  inline T max(const T &x, const T &y) {
    return x > y ? x : y;
  }
  template <class T>
  inline T min(const T &x, const T &y) {
    return x < y ? x : y;
  }
  template <class T>
  inline bool chkmax(T &x, const T &y) {
    return x < y ? x = y, true : false;
  }
  template <class T>
  inline bool chkmin(T &x, const T &y) {
    return x > y ? x = y, true : false;
  }
  inline int read() {
    register int x = 0, v = 1;
    register char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') v = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * v;
  }
  const int MOD = 1e9 + 7;
  const int MAXN = 14;
  int n;
  int a[MAXN | 1], e[MAXN | 1][MAXN | 1], f[1 << MAXN];
  int p1[1 << MAXN], p2[1 << MAXN];
  inline int add(int x, int y) { return x + y >= MOD ? x + y - MOD : x + y; }
  inline int sub(int x, int y) { return x < y ? x - y + MOD : x - y; }
  int fpow(int x, int y) {
    int res = 1, base = x;
    while (y > 0) {
      if (y & 1) res = 1ll * res * base % MOD;
      base = 1ll * base * base % MOD;
      y >>= 1;
    }
    return res;
  }
  std::pair<int, int> approx(int p, int q, int A = MOD) {
    int x = q, y = p, a = 1, b = 0;
    while (x > A) {
      std::swap(x, y);
      std::swap(a, b);
      a -= x / y * b;
      x %= y;
    }
    return std::make_pair(x, a);
  }
  int main() {
    n = read();
    for (int i = 0; i < n; ++i) a[i] = read();
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (i != j) e[i][j] = 1ll * a[i] * fpow(a[i] + a[j], MOD - 2) % MOD;
    for (int S = 1; S < 1 << n; ++S) {
      p1[S] = 1;
      for (int i = 0; i < n; ++i)
        if (S & (1 << i))
          for (int j = 0; j < n; ++j)
            if (!(S & (1 << j))) p1[S] = 1ll * p1[S] * e[i][j] % MOD;
    }
    int ans = 0;
    for (int S = 1; S < 1 << n; ++S) {
      static std::vector<int> tmp;
      tmp.clear();
      for (int T = (S - 1) & S; T > 0; T = (T - 1) & S) tmp.push_back(T);
      std::reverse(tmp.begin(), tmp.end());
      p2[0] = 1;
      f[S] = 1;
      for (int T : tmp) {
        int lb = T & (-T), id = log2(lb);
        p2[T] = p2[T ^ lb];
        for (int i = 0; i < n; ++i)
          if (!(S & (1 << i))) p2[T] = 1ll * p2[T] * e[id][i] % MOD;
        int p = 1ll * p1[T] * fpow(p2[T], MOD - 2) % MOD;
        f[S] = sub(f[S], 1ll * f[T] * p % MOD);
      }
      int p = 1;
      for (int i = 0; i < n; ++i)
        if (S & (1 << i))
          for (int j = 0; j < n; ++j)
            if (!(S & (1 << j))) p = 1ll * p * e[i][j] % MOD;
      ans = add(ans, 1ll * __builtin_popcount(S) * f[S] % MOD * p % MOD);
    }
    printf("%d\n", ans);
    return 0;
  }
  