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 INFi = 1e9 + 5;
  const int md = 1e9 + 7;
  const long long INF = 2e18;
  const int maxN = 5e5 + 1;
  double getTime() { return clock() / (double)CLOCKS_PER_SEC; }
  inline int add(const int &a, const int &b) {
    return a + b >= md ? a + b - md : a + b;
  }
  inline int sub(const int &a, const int &b) {
    return a - b < 0 ? a - b + md : a - b;
  }
  inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
  int binpow(int a, int b) {
    if (b <= 0) return 1;
    if (b % 2) return mul(a, binpow(a, b - 1));
    int m = binpow(a, b / 2);
    return mul(m, m);
  }
  int rev(int a) { return binpow(a, md - 2); }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < (n); ++i) cin >> a[i];
    vector<vector<int>> prl(n, vector<int>(1 << n));
    for (int i = 0; i < (n); ++i) {
      for (int mask = 0; mask < (1 << n); ++mask) {
        if (mask & (1 << i)) {
          prl[i][mask] = 0;
          continue;
        }
        prl[i][mask] = 1;
        for (int j = 0; j < (n); ++j) {
          if ((1 << j) & mask) {
            prl[i][mask] = mul(prl[i][mask], mul(a[i], rev(add(a[i], a[j]))));
          }
        }
      }
    }
    vector<int> pw3(n + 1);
    pw3[0] = 1;
    for (int i = 1; i <= n; ++i) pw3[i] = pw3[i - 1] * 3;
    vector<int> dp(pw3[n]);
    vector<int> prwin(pw3[n]), prlose(pw3[n]);
    vector<int> to(1 << n);
    for (int mask = 0; mask < (1 << n); ++mask) {
      for (int i = 0; i < (n); ++i) {
        if (mask & (1 << i)) to[mask] += pw3[i];
      }
    }
    for (int mask3 = 0; mask3 < (pw3[n]); ++mask3) {
      int mask1 = 0, mask2 = 0;
      {
        int c = mask3;
        for (int i = 0; i < (n); ++i) {
          int x = c % 3;
          c /= 3;
          if (x == 1) mask1 |= (1 << i);
          if (x == 2) mask2 |= (1 << i);
        }
      }
      prwin[mask3] = 1;
      prlose[mask3] = 1;
      for (int i = 0; i < (n); ++i) {
        if ((1 << i) & mask2) {
          prlose[mask3] = mul(prlose[mask3], prl[i][mask1]);
          prwin[mask3] = mul(prwin[mask3], sub(1, prl[i][mask1]));
        }
      }
    }
    for (int i = 0; i < (n); ++i) dp[pw3[i]] = 1;
    int ans = 0;
    for (int mask3 = 0; mask3 < (pw3[n]); ++mask3) {
      int mask1 = 0, mask2 = 0, mask0 = 0;
      {
        int c = mask3;
        for (int i = 0; i < (n); ++i) {
          int x = c % 3;
          c /= 3;
          if (x) mask1 |= (1 << i);
          if (x == 1) mask2 |= (1 << i);
          if (x == 0) mask0 |= (1 << i);
        }
      }
      for (int submask = mask0; submask > 0; submask = (submask - 1) & mask0) {
        int cur = dp[mask3];
        int nxt = to[mask1] * 2 + to[submask];
        cur = mul(cur, prwin[to[submask] * 2 + to[mask2]]);
        cur = mul(cur, prlose[to[submask] * 2 + to[mask1 ^ mask2]]);
        dp[nxt] = add(dp[nxt], cur);
      }
      if (!mask0) ans = add(ans, dp[mask3]);
    }
    cout << ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tests = 1;
    for (int _ = 0; _ < (tests); ++_) {
      solve();
    }
    return 0;
  }
  