#include <bits/stdc++.h>
const int N = 14;
const int Q = 1e9 + 7;
int n, a[N];
int p[N][N], g[N][1 << N];
int inverse(int x) {
  return x == 1 ? 1 : 1LL * (Q - Q / x) * inverse(Q % x) % Q;
}
int f[1 << N];
int main() {
  int T = 1;
  while (T--) {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (i != j) p[i][j] = 1LL * a[j] * inverse(a[i] + a[j]) % Q;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 1 << n; ++j) {
        g[i][j] = 1;
        for (int k = 0; k < n; ++k)
          if (j >> k & 1) g[i][j] = 1LL * g[i][j] * p[k][i] % Q;
      }
    }
    int res = 0;
    int mask = (1 << n) - 1;
    for (int i = 1; i < 1 << n; ++i) {
      int cnt = __builtin_popcount(i);
      f[i] = 1;
      int v = 1;
      for (int j = 0; j < n; ++j)
        if (i >> j & 1) v = 1LL * v * g[j][mask ^ i] % Q;
      if (cnt > 1) {
        for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
          int x = 1;
          for (int k = 0; k < n; ++k)
            if (j >> k & 1) x = 1LL * x * g[k][i ^ j] % Q;
          f[i] = (f[i] + Q - 1LL * f[j] * x % Q) % Q;
        }
      }
      res += 1LL * f[i] * v % Q * cnt % Q;
      res %= Q;
    }
    printf("%d\n", res);
  }
}

### Prompt

Please create a solution in cpp to the following problem:

<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.

### Response

```cpp
#include <bits/stdc++.h>
const int N = 14;
const int Q = 1e9 + 7;
int n, a[N];
int p[N][N], g[N][1 << N];
int inverse(int x) {
  return x == 1 ? 1 : 1LL * (Q - Q / x) * inverse(Q % x) % Q;
}
int f[1 << N];
int main() {
  int T = 1;
  while (T--) {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (i != j) p[i][j] = 1LL * a[j] * inverse(a[i] + a[j]) % Q;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 1 << n; ++j) {
        g[i][j] = 1;
        for (int k = 0; k < n; ++k)
          if (j >> k & 1) g[i][j] = 1LL * g[i][j] * p[k][i] % Q;
      }
    }
    int res = 0;
    int mask = (1 << n) - 1;
    for (int i = 1; i < 1 << n; ++i) {
      int cnt = __builtin_popcount(i);
      f[i] = 1;
      int v = 1;
      for (int j = 0; j < n; ++j)
        if (i >> j & 1) v = 1LL * v * g[j][mask ^ i] % Q;
      if (cnt > 1) {
        for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
          int x = 1;
          for (int k = 0; k < n; ++k)
            if (j >> k & 1) x = 1LL * x * g[k][i ^ j] % Q;
          f[i] = (f[i] + Q - 1LL * f[j] * x % Q) % Q;
        }
      }
      res += 1LL * f[i] * v % Q * cnt % Q;
      res %= Q;
    }
    printf("%d\n", res);
  }
}

```