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 = 1e9;
  const long long inf64 = 1e18;
  const int mod = inf + 7;
  inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
  inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
  inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
  inline void _inc(int &x, int y) {
    if ((x += y) >= mod) {
      x -= mod;
    }
  }
  inline void _dec(int &x, int y) {
    if ((x -= y) < 0) {
      x += mod;
    }
  }
  inline int _binpow(int x, int pw) {
    int res = 1;
    int tmp = x;
    while (pw > 0) {
      if (pw & 1) {
        res = _mul(res, tmp);
      }
      tmp = _mul(tmp, tmp);
      pw >>= 1;
    }
    return res;
  }
  inline int _div(int a, int b) { return _mul(a, _binpow(b, mod - 2)); }
  int slow(int n, vector<int> a) {
    vector<vector<int>> g(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        g[i][j] = _div(a[i], a[i] + a[j]);
      }
    }
    vector<pair<int, int>> edges;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) edges.emplace_back(i, j);
    }
    int m = (int)edges.size();
    int res = 0;
    for (int mask = 0; mask < (1 << m); mask++) {
      int prob = 1;
      vector<vector<int>> f(n, vector<int>(n));
      for (int i = 0; i < m; i++) {
        auto [u, v] = edges[i];
        int eprob = 0;
        if ((1 << i) & mask) {
          f[u][v] = 1;
          eprob = g[u][v];
        } else {
          f[v][u] = 1;
          eprob = g[v][u];
        }
        prob = _mul(prob, eprob);
      }
      for (int k = 0; k < n; k++) {
        for (int i = 0; i < n; i++) {
          for (int j = 0; j < n; j++) {
            f[i][j] |= f[i][k] & f[k][j];
          }
        }
      }
      for (int v = 0; v < n; v++) f[v][v] = 1;
      int sz = 0;
      for (int v = 0; v < n; v++) {
        int ok = 1;
        for (int u = 0; u < n; u++) ok &= f[v][u];
        sz += ok;
      }
      _inc(res, _mul(sz, prob));
    }
    return res;
  }
  int fast(int n, vector<int> a) {
    vector<vector<int>> g(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        g[i][j] = _div(a[i], a[i] + a[j]);
      }
    }
    vector<int> comp2(1 << n);
    for (int mask = 0; mask < (1 << n); mask++) {
      vector<int> in;
      for (int i = 0; i < n; i++) {
        if ((1 << i) & mask) in.push_back(i);
      }
      for (int A = mask; A > 0; A = (A - 1) & mask) {
        int B = mask ^ A;
        if (!A || !B) continue;
        int prob = 1;
        for (int u : in) {
          if ((1 << u) & B) continue;
          for (int v : in) {
            if ((1 << v) & A) continue;
            prob = _mul(prob, g[u][v]);
          }
        }
        _inc(comp2[mask], _mul(prob, comp2[A]));
      }
      comp2[mask] = _sub(1, comp2[mask]);
    }
    int res = 0;
    for (int mask = 0; mask < (1 << n); mask++) {
      vector<int> in, out;
      for (int i = 0; i < n; i++) {
        if ((1 << i) & mask) {
          in.push_back(i);
        } else {
          out.push_back(i);
        }
      }
      int coef = 1;
      for (int u : in) {
        for (int v : out) {
          coef = _mul(coef, g[u][v]);
        }
      }
      int prob = _mul(coef, comp2[mask]);
      _inc(res, _mul(prob, (int)in.size()));
    }
    return res;
  }
  void work() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    int res = fast(n, a);
    cout << res << "\n";
  }
  void test() {
    mt19937 rnd(42);
    const int MAXN = 5;
    const int MAXX = 10;
    while (1) {
      int n = rnd() % MAXN + 1;
      vector<int> a(n);
      for (int i = 0; i < n; i++) a[i] = rnd() % MAXX + 1;
      int fs = fast(n, a);
      int sl = slow(n, a);
      if (fs == sl) {
        cout << "OK(" << fs << ")" << endl;
      } else {
        cout << "WA\n";
        cout << "exp = " << sl << "\n";
        cout << "fnd = " << fs << "\n";
        cout << n << "\n";
        for (int i = 0; i < n; i++) cout << a[i] << " ";
        cout << "\n";
        break;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    work();
    return 0;
  }
  