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 target("avx,avx2,fma")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  using ll = long long;
  using db = long double;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<db, db>;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vs = vector<string>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
  const int MAXN = 3e6 + 5;
  const int MOD = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  const ll INF64 = ll(4e18) + 5;
  const db EPS = 1e-9;
  const db PI = acos((db)-1);
  int add(int a, int b) {
    a += b;
    if (a >= MOD) a -= MOD;
    if (a < 0) a += MOD;
    return a;
  }
  int mul(int a, int b) {
    int ret = (1LL * a * b) % MOD;
    return ret;
  }
  int pw(int a, int n) {
    int ret = 1;
    while (n > 0) {
      if (n & 1 == 1) {
        ret = mul(ret, a);
      }
      a = mul(a, a);
      n = n >> 1;
    }
    return ret;
  }
  int inv(int a) { return pw(a, MOD - 2); }
  int divide(int a, int b) { return mul(a, inv(b)); }
  void solve() {
    int n;
    cin >> n;
    vi ina(MAXN);
    for (int i = 1; i < MAXN; i++) {
      ina[i] = inv(i);
    }
    vi a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int ans = 0;
    int m = n - 1;
    int dp[(1 << m)];
    int h[(1 << m)][m];
    for (int i = 0; i < n; i++) {
      for (int mask = 0; mask < (1 << m); mask++) {
        dp[mask] = 0;
      }
      vi b(n - 1);
      int at = 0;
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        b[at] = a[j];
        at++;
      }
      int ai = a[i];
      at = 0;
      int k = 1;
      for (int mask = 0; mask < (1 << m); mask++) {
        for (int j = 0; j < m; j++) {
          int val = mask & (1 << j);
          if (val || mask == 0) continue;
          h[mask][j] = 1;
          for (int k = 0; k < m; k++) {
            if (mask & (1 << k)) {
              h[mask][j] = mul(h[mask][j], mul(b[j], ina[b[j] + b[k]]));
            }
          }
        }
      }
      for (int mask = (1 << m) - 1; mask >= 0; mask--) {
        dp[mask] = 1;
        int cnt = 0;
        for (int j = 0; j < m; j++) {
          int val = mask & (1 << j);
          if (val == 0) continue;
          cnt++;
          for (int k = 0; k < m; k++) {
            if (mask & (1 << k)) continue;
            int aux = mul(b[j], ina[b[j] + b[k]]);
            dp[mask] = mul(dp[mask], aux);
          }
          int aux = mul(b[j], ina[ai + b[j]]);
          dp[mask] = mul(dp[mask], aux);
        }
        for (int sub = mask; sub < (1 << m); sub = (sub + 1) | mask) {
          if (sub == mask) continue;
          int x = dp[sub];
          int com = 0;
          for (int j = 0; j < m; j++) {
            int v1 = mask & (1 << j);
            int v2 = sub & (1 << j);
            if (v2 && v1 == 0) {
              com |= (1 << j);
            }
          }
          for (int j = 0; j < m; j++) {
            int v1 = mask & (1 << j);
            int v2 = sub & (1 << j);
            if (v1) {
              x = mul(x, h[com][j]);
            }
          }
          dp[mask] = add(dp[mask], -x);
        }
      }
      ans = add(ans, dp[0]);
    }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  