#include <bits/stdc++.h>
using namespace std;
int n, ans;
int a[19];
int inv[2000009];
int f[17009];
int tag[19];
int qpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = (long long)a * a % 1000000007)
    if (b & 1) ans = (long long)ans * a % 1000000007;
  return ans;
}
void init() {
  for (int i = 1; i <= 2000000; i++) inv[i] = qpow(i, 1000000007 - 2);
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int s = 1; s < (1 << n); s++) {
    int sm = 0;
    for (int t = (s - 1) & s; t; t = (t - 1) & s) {
      int p = s ^ t, pb = 1;
      for (int i = 1; i <= n; i++)
        if (t >> (i - 1) & 1)
          for (int j = 1; j <= n; j++)
            if (p >> (j - 1) & 1)
              pb = (long long)pb * a[i] % 1000000007 * inv[a[i] + a[j]] %
                   1000000007;
      sm = (sm + (long long)f[t] * pb) % 1000000007;
    }
    f[s] = (1000000007 + 1 - sm) % 1000000007;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int s = 1; s < (1 << n); s++)
      if (s >> (i - 1) & 1) {
        int pb = 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))
                pb = (long long)pb * a[i] % 1000000007 * inv[a[i] + a[j]] %
                     1000000007;
        ans = (ans + (long long)f[s] * pb) % 1000000007;
      }
  printf("%d", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

<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;
int n, ans;
int a[19];
int inv[2000009];
int f[17009];
int tag[19];
int qpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = (long long)a * a % 1000000007)
    if (b & 1) ans = (long long)ans * a % 1000000007;
  return ans;
}
void init() {
  for (int i = 1; i <= 2000000; i++) inv[i] = qpow(i, 1000000007 - 2);
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int s = 1; s < (1 << n); s++) {
    int sm = 0;
    for (int t = (s - 1) & s; t; t = (t - 1) & s) {
      int p = s ^ t, pb = 1;
      for (int i = 1; i <= n; i++)
        if (t >> (i - 1) & 1)
          for (int j = 1; j <= n; j++)
            if (p >> (j - 1) & 1)
              pb = (long long)pb * a[i] % 1000000007 * inv[a[i] + a[j]] %
                   1000000007;
      sm = (sm + (long long)f[t] * pb) % 1000000007;
    }
    f[s] = (1000000007 + 1 - sm) % 1000000007;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int s = 1; s < (1 << n); s++)
      if (s >> (i - 1) & 1) {
        int pb = 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))
                pb = (long long)pb * a[i] % 1000000007 * inv[a[i] + a[j]] %
                     1000000007;
        ans = (ans + (long long)f[s] * pb) % 1000000007;
      }
  printf("%d", ans);
  return 0;
}

```