#include <bits/stdc++.h>
using namespace std;
const int md = 1000000007;
int a[15], p[15][15], f[100001], u[15], v[15];
int pw(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % md)
    if (y & 1) ans = 1ll * ans * x % md;
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      p[i][j] = 1ll * a[i] * pw(a[i] + a[j], md - 2) % md;
  int ans = 0;
  for (int s = 1; s < 1 << n; s++) {
    f[s] = 1;
    int oo = 1;
    for (int i = 1; i <= n; i++)
      if (s >> i - 1 & 1)
        for (int j = 1; j <= n; j++)
          if (!(s >> j - 1 & 1)) oo = 1ll * oo * p[i][j] % md;
    for (int s2 = s - 1 & s; s2 != 0; s2 = s2 - 1 & s) {
      int o = 1;
      u[0] = v[0] = 0;
      for (int i = 1; i <= n; i++)
        if (s >> i - 1 & 1)
          if (s2 >> i - 1 & 1)
            u[++u[0]] = i;
          else
            v[++v[0]] = i;
      for (int i = 1; i <= u[0]; i++)
        for (int j = 1; j <= v[0]; j++) o = 1ll * o * p[u[i]][v[j]] % md;
      f[s] = ((f[s] - 1ll * o * f[s2]) % md + md) % md;
    }
    int num = 0;
    for (int i = 1; i <= n; i++)
      if (s >> i - 1 & 1) num++;
    ans = (ans + 1ll * f[s] * oo % md * num) % md;
  }
  printf("%d\n", ans);
}

### 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>
using namespace std;
const int md = 1000000007;
int a[15], p[15][15], f[100001], u[15], v[15];
int pw(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % md)
    if (y & 1) ans = 1ll * ans * x % md;
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      p[i][j] = 1ll * a[i] * pw(a[i] + a[j], md - 2) % md;
  int ans = 0;
  for (int s = 1; s < 1 << n; s++) {
    f[s] = 1;
    int oo = 1;
    for (int i = 1; i <= n; i++)
      if (s >> i - 1 & 1)
        for (int j = 1; j <= n; j++)
          if (!(s >> j - 1 & 1)) oo = 1ll * oo * p[i][j] % md;
    for (int s2 = s - 1 & s; s2 != 0; s2 = s2 - 1 & s) {
      int o = 1;
      u[0] = v[0] = 0;
      for (int i = 1; i <= n; i++)
        if (s >> i - 1 & 1)
          if (s2 >> i - 1 & 1)
            u[++u[0]] = i;
          else
            v[++v[0]] = i;
      for (int i = 1; i <= u[0]; i++)
        for (int j = 1; j <= v[0]; j++) o = 1ll * o * p[u[i]][v[j]] % md;
      f[s] = ((f[s] - 1ll * o * f[s2]) % md + md) % md;
    }
    int num = 0;
    for (int i = 1; i <= n; i++)
      if (s >> i - 1 & 1) num++;
    ans = (ans + 1ll * f[s] * oo % md * num) % md;
  }
  printf("%d\n", ans);
}

```