#include <bits/stdc++.h>
using namespace std;
const int N = 14;
const long long mod = 1e9 + 7;
int n;
long long a[N], w[N][N], f[1 << N];
long long ans;
long long qpow(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return r;
}
long long rev(long long x) { return qpow(x, mod - 2); }
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      w[i][j] = a[i] * rev(a[i] + a[j]) % mod;
    }
  for (int S = 1; S < (1 << n); ++S) {
    f[S] = 1;
    for (int T = (S - 1) & S; T != S; T = (T - 1) & S) {
      long long val = f[T];
      for (int i = 0; i < n; ++i)
        if (T & 1 << i) {
          for (int j = 0; j < n; ++j)
            if ((S ^ T) & 1 << j) {
              val *= w[i][j];
              val %= mod;
            }
        }
      f[S] = (f[S] - val + mod) % mod;
    }
    long long val = f[S];
    for (int i = 0; i < n; ++i)
      if (S & 1 << i) {
        for (int j = 0; j < n; ++j)
          if (!(S & 1 << j)) {
            val *= w[i][j];
            val %= mod;
          }
      }
    ans += val * __builtin_popcount(S) % mod;
  }
  cout << ans % mod << endl;
}

### Prompt

Generate 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 N = 14;
const long long mod = 1e9 + 7;
int n;
long long a[N], w[N][N], f[1 << N];
long long ans;
long long qpow(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return r;
}
long long rev(long long x) { return qpow(x, mod - 2); }
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      w[i][j] = a[i] * rev(a[i] + a[j]) % mod;
    }
  for (int S = 1; S < (1 << n); ++S) {
    f[S] = 1;
    for (int T = (S - 1) & S; T != S; T = (T - 1) & S) {
      long long val = f[T];
      for (int i = 0; i < n; ++i)
        if (T & 1 << i) {
          for (int j = 0; j < n; ++j)
            if ((S ^ T) & 1 << j) {
              val *= w[i][j];
              val %= mod;
            }
        }
      f[S] = (f[S] - val + mod) % mod;
    }
    long long val = f[S];
    for (int i = 0; i < n; ++i)
      if (S & 1 << i) {
        for (int j = 0; j < n; ++j)
          if (!(S & 1 << j)) {
            val *= w[i][j];
            val %= mod;
          }
      }
    ans += val * __builtin_popcount(S) % mod;
  }
  cout << ans % mod << endl;
}

```