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;
  using ll = long long;
  using pii = pair<int, int>;
  using pil = pair<int, ll>;
  using pli = pair<ll, int>;
  using pll = pair<ll, ll>;
  const int MOD = 1000000007;
  const int inf = (1 << 30) - 1;
  const ll INF = (1LL << 60) - 1;
  template <typename T>
  bool chmax(T &x, const T &y) {
    return (x < y) ? (x = y, true) : false;
  };
  template <typename T>
  bool chmin(T &x, const T &y) {
    return (x > y) ? (x = y, true) : false;
  };
  struct io_setup {
    io_setup() {
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
      cout << fixed << setprecision(15);
    }
  } io_setup;
  template <int mod>
  struct Mod_Int {
    int x;
    Mod_Int() : x(0) {}
    Mod_Int(long long y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    static int get_mod() { return mod; }
    Mod_Int &operator+=(const Mod_Int &p) {
      if ((x += p.x) >= mod) x -= mod;
      return *this;
    }
    Mod_Int &operator-=(const Mod_Int &p) {
      if ((x += mod - p.x) >= mod) x -= mod;
      return *this;
    }
    Mod_Int &operator*=(const Mod_Int &p) {
      x = (int)(1LL * x * p.x % mod);
      return *this;
    }
    Mod_Int &operator/=(const Mod_Int &p) {
      *this *= p.inverse();
      return *this;
    }
    Mod_Int &operator++() { return *this += Mod_Int(1); }
    Mod_Int operator++(int) {
      Mod_Int tmp = *this;
      ++*this;
      return tmp;
    }
    Mod_Int &operator--() { return *this -= Mod_Int(1); }
    Mod_Int operator--(int) {
      Mod_Int tmp = *this;
      --*this;
      return tmp;
    }
    Mod_Int operator-() const { return Mod_Int(-x); }
    Mod_Int operator+(const Mod_Int &p) const { return Mod_Int(*this) += p; }
    Mod_Int operator-(const Mod_Int &p) const { return Mod_Int(*this) -= p; }
    Mod_Int operator*(const Mod_Int &p) const { return Mod_Int(*this) *= p; }
    Mod_Int operator/(const Mod_Int &p) const { return Mod_Int(*this) /= p; }
    bool operator==(const Mod_Int &p) const { return x == p.x; }
    bool operator!=(const Mod_Int &p) const { return x != p.x; }
    Mod_Int inverse() const {
      assert(*this != Mod_Int(0));
      return pow(mod - 2);
    }
    Mod_Int pow(long long k) const {
      Mod_Int now = *this, ret = 1;
      for (; k > 0; k >>= 1, now *= now) {
        if (k & 1) ret *= now;
      }
      return ret;
    }
    friend ostream &operator<<(ostream &os, const Mod_Int &p) {
      return os << p.x;
    }
    friend istream &operator>>(istream &is, Mod_Int &p) {
      long long a;
      is >> a;
      p = Mod_Int<mod>(a);
      return is;
    }
  };
  using mint = Mod_Int<MOD>;
  int main() {
    int N;
    cin >> N;
    vector<mint> a(N);
    for (int i = 0; i < N; i++) cin >> a[i];
    vector<vector<mint>> iv(N, vector<mint>(N));
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        iv[i][j] = (a[i] + a[j]).inverse();
      }
    }
    vector<vector<mint>> dp1(N, vector<mint>(1 << N, 1));
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < 1 << N; j++) {
        for (int k = 0; k < N; k++) {
          if ((j >> k) & 1) {
            dp1[i][j] *= a[i] * iv[i][k];
          }
        }
      }
    }
    vector<mint> dp(1 << N, 1);
    for (int i = 0; i < 1 << N; i++) {
      for (int j = i; j > 0; j--, j &= i) {
        if (j == i) continue;
        mint tmp = 1;
        for (int k = 0; k < N; k++) {
          if ((j >> k) & 1) {
            tmp *= dp1[k][i & ~j];
          }
        }
        dp[i] -= dp[j] * tmp;
      }
    }
    mint ans = 0;
    for (int i = 0; i < 1 << N; i++) {
      if (i == 0) continue;
      int j = ((1 << N) - 1) ^ i;
      for (int k = 0; k < N; k++) {
        if ((i >> k) & 1) dp[i] *= dp1[k][j];
      }
      ans += dp[i] * __builtin_popcount(i);
    }
    cout << ans << '\n';
  }
  