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 long long MOD = 1e9 + 7;
  const long long N = 16;
  const long long M = 5e6 + 5;
  long long n, a[N], P[20005], DF[M], res = 0, Div[N][N], MaxM;
  vector<pair<long long, long long> > lst;
  void add(long long &a, long long b) { a = (a + b) % MOD; }
  long long Pow(long long a, long long b) {
    if (b == 0) return 1;
    long long t = Pow(a, b / 2);
    if (b % 2 == 0) return t * t % MOD;
    return t * t % MOD * a % MOD;
  }
  long long check_bit(long long i, long long mask) {
    return (mask & (1 << i)) > 0;
  }
  long long Find(pair<long long, long long> t) {
    long long low = 0, high = lst.size() - 1;
    while (low <= high) {
      long long mid = (low + high) / 2;
      if (lst[mid].first < t.first)
        low = mid + 1;
      else {
        if (lst[mid].first > t.first)
          high = mid - 1;
        else {
          if (lst[mid].second > t.second)
            high = mid - 1;
          else {
            if (lst[mid].second < t.second)
              low = mid + 1;
            else
              return mid;
          }
        }
      }
    }
  }
  void Prepare() {
    for (long long i = 0; i <= n - 1; i++)
      for (long long j = i + 1; j <= n - 1; j++) {
        if (i == j) continue;
        Div[i][j] = Pow(a[i] + a[j], MOD - 2) % MOD;
        Div[j][i] = Div[i][j];
      }
    for (long long mask1 = 1; mask1 <= MaxM; mask1++) {
      for (long long mask2 = 0; mask2 <= (MaxM ^ mask1); mask2++) {
        if ((mask1 & mask2) != 0) continue;
        lst.push_back({mask1, mask2});
        long long num = __builtin_popcount(mask1),
                  firstPos = __builtin_ctz(mask1);
        long long cur = lst.size() - 1;
        DF[cur] = 1;
        if (num == 1) {
          for (long long i = 0; i < n; i++) {
            if (check_bit(i, mask2)) {
              DF[cur] = DF[cur] * a[firstPos] % MOD * Div[firstPos][i] % MOD;
            }
          }
        } else {
          long long subMask = (1 << firstPos), ano_subMask = (mask1 ^ subMask);
          DF[cur] =
              DF[Find({subMask, mask2})] * DF[Find({ano_subMask, mask2})] % MOD;
        }
      }
    }
  }
  void Solve() {
    long long res = 0;
    for (long long mask = 1; mask <= MaxM; mask++) {
      long long num_Win = __builtin_popcount(mask);
      long long Lose = (MaxM ^ mask);
      P[mask] = 1;
      for (long long subMask = 1; subMask <= mask - 1; subMask++) {
        if ((mask & subMask) != subMask) continue;
        long long ano_subMask = (mask ^ subMask);
        P[mask] = (P[mask] - P[subMask] * DF[Find({subMask, ano_subMask})] % MOD +
                   MOD) %
                  MOD;
      }
      add(res, P[mask] * DF[Find({mask, Lose})] % MOD * num_Win % MOD);
    }
    cout << res << '\n';
  }
  void in() {
    cin >> n;
    for (long long i = 0; i <= n - 1; i++) cin >> a[i];
    MaxM = (1 << n) - 1;
  }
  void process() {
    Prepare();
    Solve();
  }
  int main() {
    if (fopen("trash.inp", "r"))
      freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    in();
    process();
  }
  