#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 2e6 + 10;
int power(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = 1LL * res * a % mod;
    }
    a = 1LL * a * a % mod;
    b /= 2;
  }
  return res;
}
int main() {
  int test = 1, a[15], p[1 << 15], pp[15][15];
  while (test--) {
    int n, ans = 0;
    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++) {
        pp[i][j] = 1LL * a[i] * power(a[i] + a[j], mod - 2) % mod;
      }
    }
    p[0] = 0;
    for (int i = 1; i < (1 << n); i++) {
      p[i] = 1;
      for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
        int g = p[j];
        for (int k = 0; k < n; k++) {
          if (j & (1 << k)) {
            for (int l = 0; l < n; l++) {
              if (j & (1 << l)) {
                continue;
              }
              if (i & (1 << l)) {
                g = 1LL * g * pp[k][l] % mod;
              }
            }
          }
        }
        p[i] -= g % mod;
        if (p[i] < 0) {
          p[i] += mod;
        }
      }
      int g = 1;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          for (int k = 0; k < n; k++) {
            if (i & (1 << k)) {
              continue;
            }
            g = 1LL * g * pp[j][k] % mod;
          }
        }
      }
      ans += 1LL * p[i] * g % mod * __builtin_popcount(i) % mod;
      ans %= mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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 mod = 1e9 + 7;
const int N = 2e6 + 10;
int power(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = 1LL * res * a % mod;
    }
    a = 1LL * a * a % mod;
    b /= 2;
  }
  return res;
}
int main() {
  int test = 1, a[15], p[1 << 15], pp[15][15];
  while (test--) {
    int n, ans = 0;
    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++) {
        pp[i][j] = 1LL * a[i] * power(a[i] + a[j], mod - 2) % mod;
      }
    }
    p[0] = 0;
    for (int i = 1; i < (1 << n); i++) {
      p[i] = 1;
      for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
        int g = p[j];
        for (int k = 0; k < n; k++) {
          if (j & (1 << k)) {
            for (int l = 0; l < n; l++) {
              if (j & (1 << l)) {
                continue;
              }
              if (i & (1 << l)) {
                g = 1LL * g * pp[k][l] % mod;
              }
            }
          }
        }
        p[i] -= g % mod;
        if (p[i] < 0) {
          p[i] += mod;
        }
      }
      int g = 1;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          for (int k = 0; k < n; k++) {
            if (i & (1 << k)) {
              continue;
            }
            g = 1LL * g * pp[j][k] % mod;
          }
        }
      }
      ans += 1LL * p[i] * g % mod * __builtin_popcount(i) % mod;
      ans %= mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```