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;
  const int inf = (int)1.01e9;
  const long long infll = (long long)1.01e18;
  const long double eps = 1e-9;
  const long double pi = acos((long double)-1);
  mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
  int rnd(int x) { return mrand() % x; }
  void precalc() {}
  const int mod = (int)1e9 + 7;
  int mul(int a, int b) { return (long long)a * b % mod; }
  void add(int &a, int b) {
    a += b;
    if (a >= mod) {
      a -= mod;
    }
  }
  pair<int, int> gcd(int a, int b) {
    if (!b) {
      return make_pair(1, 0);
    }
    auto p = gcd(b, a % b);
    return make_pair(p.second, p.first - (a / b) * p.second);
  }
  int inv(int x) {
    int res = gcd(x, mod).first % mod;
    if (res < 0) {
      res += mod;
    }
    return res;
  }
  const int maxn = 15, pown = (1 << maxn);
  int n;
  int a[maxn];
  bool read() {
    if (scanf("%d", &n) < 1) {
      return false;
    }
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    return true;
  }
  int g[maxn][maxn];
  int mem[maxn][pown];
  int dp[pown];
  void solve() {
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        int q = inv(a[i] + a[j]);
        g[i][j] = mul(a[i], q);
        g[j][i] = mul(a[j], q);
      }
    }
    for (int v = 0; v < n; ++v) {
      for (int msk = 0; msk < (1 << n); ++msk) {
        auto &cur = mem[v][msk];
        cur = 1;
        for (int u = 0; u < n; ++u) {
          if (!(msk & (1 << u))) {
            continue;
          }
          cur = mul(cur, g[v][u]);
        }
      }
    }
    for (int msk = 1; msk < (1 << n); ++msk) {
      auto &cur = dp[msk];
      cur = 1;
      for (int sub = ((msk - 1) & msk); sub > 0; sub = ((sub - 1) & msk)) {
        int pr = dp[sub];
        for (int v = 0; v < n; ++v) {
          if (!(sub & (1 << v))) {
            continue;
          }
          pr = mul(pr, mem[v][msk ^ sub]);
        }
        add(cur, mod - pr);
      }
    }
    int res = 0;
    for (int msk = 1; msk < (1 << n); ++msk) {
      auto cur = dp[msk];
      for (int v = 0; v < n; ++v) {
        if (!(msk & (1 << v))) {
          continue;
        }
        cur = mul(cur, mem[v][((1 << n) - 1) ^ msk]);
      }
      add(res, mul(__builtin_popcount(msk), cur));
    }
    printf("%d\n", res);
  }
  int main() {
    precalc();
    while (read()) {
      solve();
    }
    return 0;
  }
  