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 N1 = 5e6 + 5;
  const long long N2 = 2e4 + 5;
  long long n, a[16], Up[16][16], WL[N1], P[N2], Res, Lim;
  vector<pair<long long, long long> > Match;
  void Input() {
    cin >> n;
    for (long long i = 0; i < n; ++i) cin >> a[i];
  }
  long long Pow(long long A, long long B) {
    if (B == 0) return 1;
    long long C = Pow(A, B / 2);
    if (B % 2 == 0) return C * C % MOD;
    return C * C % MOD * A % MOD;
  }
  long long Get_Pos(pair<long long, long long> A) {
    long long Low = 0, High = Match.size();
    High--;
    while (Low <= High) {
      long long Mid = (Low + High) / 2;
      if (Match[Mid].first < A.first)
        Low = Mid + 1;
      else {
        if (Match[Mid].first > A.first)
          High = Mid - 1;
        else {
          if (Match[Mid].second < A.second)
            Low = Mid + 1;
          else {
            if (Match[Mid].second > A.second)
              High = Mid - 1;
            else
              return Mid + 1;
          }
        }
      }
    }
  }
  bool In_Mask(long long x, long long Mask) { return (Mask & (1ll << x)) > 0; }
  void Cal_WinToLose() {
    for (long long i = 0; i < n; ++i) {
      for (long long j = i + 1; j < n; ++j) {
        Up[i][j] = Pow(a[i] + a[j], MOD - 2);
        Up[j][i] = Up[i][j];
      }
    }
    for (long long Mask1 = 1; Mask1 <= Lim; ++Mask1) {
      for (long long Mask2 = 0; Mask2 <= (Lim ^ Mask1); ++Mask2) {
        if ((Mask1 & Mask2) != 0) continue;
        Match.push_back({Mask1, Mask2});
        long long Cnt = Match.size();
        long long NumBit = __builtin_popcount(Mask1);
        long long PosBit = __builtin_ctz(Mask1);
        WL[Cnt] = 1;
        if (NumBit == 1) {
          for (long long i = 0; i < n; ++i) {
            if (In_Mask(i, Mask2))
              WL[Cnt] = WL[Cnt] * a[PosBit] % MOD * Up[PosBit][i] % MOD;
          }
        } else {
          long long subMask1 = (1ll << PosBit), subMask2 = (Mask1 ^ subMask1);
          WL[Cnt] = WL[Get_Pos({subMask1, Mask2})] *
                    WL[Get_Pos({subMask2, Mask2})] % MOD;
        }
      }
    }
  }
  void Cal_Probability() {
    for (long long Mask = 1; Mask <= Lim; ++Mask) {
      P[Mask] = 1;
      long long NumWin = __builtin_popcount(Mask);
      long long Lose = (Lim ^ Mask);
      for (long long subMask = 1; subMask < Mask; ++subMask) {
        if ((Mask | subMask) != Mask) continue;
        long long exMask = (Mask ^ subMask);
        P[Mask] =
            (P[Mask] - P[subMask] * WL[Get_Pos({subMask, exMask})] % MOD + MOD) %
            MOD;
      }
      Res = (Res + NumWin * (P[Mask] * WL[Get_Pos({Mask, Lose})] % MOD)) % MOD;
    }
  }
  void Solve() {
    Lim = (1ll << n) - 1;
    Res = 0;
    Cal_WinToLose();
    Cal_Probability();
    cout << Res;
  }
  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);
    Input();
    Solve();
  }
  