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;
  namespace IO {
  const int RLEN = 1 << 22 | 1;
  char ibuf[RLEN], *ib, *ob;
  inline char gc() {
    (ib == ob) && (ob = (ib = ibuf) + fread(ibuf, 1, RLEN, stdin));
    return (ib == ob) ? EOF : *ib++;
  }
  inline int read() {
    char ch = gc();
    int res = 0;
    bool f = 1;
    while (!isdigit(ch)) f ^= ch == '-', ch = gc();
    while (isdigit(ch)) res = (res * 10) + (ch ^ 48), ch = gc();
    return f ? res : -res;
  }
  inline long long readll() {
    char ch = gc();
    long long res = 0;
    bool f = 1;
    while (!isdigit(ch)) f ^= ch == '-', ch = gc();
    while (isdigit(ch)) res = (res * 10) + (ch ^ 48), ch = gc();
    return f ? res : -res;
  }
  inline char readchar() {
    char ch = gc();
    while (isspace(ch)) ch = gc();
    return ch;
  }
  inline int readstring(char *s) {
    int top = 0;
    char ch = gc();
    while (isspace(ch)) ch = gc();
    while (!isspace(ch) && ch != EOF) s[++top] = ch, ch = gc();
    s[top + 1] = '\0';
    return top;
  }
  }  // namespace IO
  using IO::read;
  using IO::readchar;
  using IO::readll;
  using IO::readstring;
  template <typename tp>
  inline bool chemx(tp &a, tp b) {
    return (a < b) ? (a = b, 1) : 0;
  }
  template <typename tp>
  inline bool chemn(tp &a, tp b) {
    return (a > b) ? (a = b, 1) : 0;
  }
  const int mod = 1e9 + 7;
  inline int add(int a, int b) {
    return (a + b) >= mod ? (a + b - mod) : (a + b);
  }
  inline int dec(int a, int b) { return (a < b) ? (a - b + mod) : (a - b); }
  inline int mul(int a, int b) {
    static long long r;
    r = (long long)a * b;
    return (r >= mod) ? (r % mod) : r;
  }
  inline void Add(int &a, int b) { a = (a + b) >= mod ? (a + b - mod) : (a + b); }
  inline void Dec(int &a, int b) { a = (a < b) ? (a - b + mod) : (a - b); }
  inline void Mul(int &a, int b) {
    static long long r;
    r = (long long)a * b;
    a = (r >= mod) ? (r % mod) : r;
  }
  inline int ksm(int a, int b, int res = 1) {
    for (; b; b >>= 1, Mul(a, a)) (b & 1) && (Mul(res, a), 1);
    return res;
  }
  inline int Inv(int x) { return ksm(x, mod - 2); }
  inline int fix(long long x) {
    x %= mod;
    return (x < 0) ? x + mod : x;
  }
  const int N = 15, M = (1 << 15) | 5;
  int n, a[N], pw[N][N], p1[N][M], f[M], id[M], lim;
  inline int lb(int x) { return x & (-x); }
  int calc(int s) {}
  int main() {
    n = read(), lim = 1 << n;
    for (int i = 1; i <= n; i++)
      a[i] = read(), id[1 << (i - 1)] = i, p1[i][0] = 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (i ^ j) pw[i][j] = mul(a[i], Inv(a[i] + a[j]));
    for (int s = 1; s < lim; s++)
      for (int i = 1; i <= n; i++)
        if (!(s & (1 << (i - 1))))
          p1[i][s] = mul(p1[i][s - lb(s)], pw[id[lb(s)]][i]);
    f[0] = 1;
    int res = 0;
    for (int s = 1; s < lim; s++) {
      int pt = 1;
      for (int ss = s; ss; ss = (ss - 1) & s)
        if (ss != s) {
          int ret = 1;
          for (int i = 1; i <= n; i++)
            if (!(s & (1 << (i - 1)))) Mul(ret, p1[i][ss]);
          Dec(pt, mul(f[ss], Inv(ret)));
        }
      int pc = 1;
      for (int i = 1; i <= n; i++)
        if (!(s & (1 << (i - 1)))) Mul(pc, p1[i][s]);
      f[s] = mul(pt, pc);
      Add(res, mul(f[s], __builtin_popcount(s)));
    }
    cout << res << '\n';
    return 0;
  }
  