#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, a[20], f[40005], ans, s[105][105];
int P(int x, int y) {
  int p = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (x & (1 << i - 1))
        if (y & (1 << j - 1)) p = 1ll * p * s[i][j] % mod;
  return p;
}
int Power(int x, int y) {
  int r = 1;
  while (y) {
    if (y & 1) r = 1ll * r * x % mod;
    x = 1ll * x * x % mod, y >>= 1;
  }
  return r;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      s[i][j] = 1ll * a[i] * Power(a[i] + a[j], mod - 2) % mod;
  for (int i = 1; i < (1 << n); i++) {
    f[i] = 1;
    for (int j = i & (i - 1); j; j = (j - 1) & i) {
      f[i] = (f[i] - 1ll * f[j] * P(j, i - j) % mod + mod) % mod;
    }
  }
  for (int i = 1; i < (1 << n); i++)
    ans = (ans +
           1ll * f[i] * P(i, (1 << n) - 1 - i) % mod * __builtin_popcount(i)) %
          mod;
  cout << 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 mod = 1e9 + 7;
int n, a[20], f[40005], ans, s[105][105];
int P(int x, int y) {
  int p = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (x & (1 << i - 1))
        if (y & (1 << j - 1)) p = 1ll * p * s[i][j] % mod;
  return p;
}
int Power(int x, int y) {
  int r = 1;
  while (y) {
    if (y & 1) r = 1ll * r * x % mod;
    x = 1ll * x * x % mod, y >>= 1;
  }
  return r;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      s[i][j] = 1ll * a[i] * Power(a[i] + a[j], mod - 2) % mod;
  for (int i = 1; i < (1 << n); i++) {
    f[i] = 1;
    for (int j = i & (i - 1); j; j = (j - 1) & i) {
      f[i] = (f[i] - 1ll * f[j] * P(j, i - j) % mod + mod) % mod;
    }
  }
  for (int i = 1; i < (1 << n); i++)
    ans = (ans +
           1ll * f[i] * P(i, (1 << n) - 1 - i) % mod * __builtin_popcount(i)) %
          mod;
  cout << ans;
}

```