#include <bits/stdc++.h>
using namespace std;
const int N = 14, MOD = 1e9 + 7;
int n;
long long a[N];
long long p[N][N];
long long H[N][1 << N];
long long f[1 << N];
long long g[1 << N];
long long pow_mod(long long x, int n) {
  long long y = 1;
  while (n) {
    if (n & 1) y = (long long)y * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return y;
}
int main() {
  cin.tie(0), cout.tie(0);
  ios::sync_with_stdio(false);
  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++)
      p[i][j] = a[i] * pow_mod(a[i] + a[j], MOD - 2) % MOD;
  int all = (1 << n) - 1;
  for (int i = 0; i < n; i++) {
    H[i][0] = 1;
    for (int j = 0; j < n; j++) H[i][1 << j] = p[i][j];
    for (int S = 1; S <= all; S++)
      H[i][S] = H[i][(S - 1) & S] * H[i][-S & S] % MOD;
  }
  int ans = 0;
  for (int S = 1; S < 1 << n; S++) {
    g[0] = 1;
    for (int tS = (S - 1) & S; tS; tS = (tS - 1) & S) {
      int mask = S ^ tS;
      int x = __builtin_ffs(mask) - 1;
      g[mask] = g[mask ^ (1 << x)] * H[x][all ^ S] % MOD;
      f[S] = (f[S] - f[tS] * g[mask]) % MOD;
    }
    int x = __builtin_ffs(S) - 1;
    g[S] = g[S ^ (1 << x)] * H[x][all ^ S] % MOD;
    f[S] = (f[S] + g[S]) % MOD;
    ans = (ans + f[S] * __builtin_popcount(S)) % MOD;
  }
  cout << (ans + MOD) % MOD << "\n";
}

### Prompt

Create a solution in CPP for 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, MOD = 1e9 + 7;
int n;
long long a[N];
long long p[N][N];
long long H[N][1 << N];
long long f[1 << N];
long long g[1 << N];
long long pow_mod(long long x, int n) {
  long long y = 1;
  while (n) {
    if (n & 1) y = (long long)y * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return y;
}
int main() {
  cin.tie(0), cout.tie(0);
  ios::sync_with_stdio(false);
  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++)
      p[i][j] = a[i] * pow_mod(a[i] + a[j], MOD - 2) % MOD;
  int all = (1 << n) - 1;
  for (int i = 0; i < n; i++) {
    H[i][0] = 1;
    for (int j = 0; j < n; j++) H[i][1 << j] = p[i][j];
    for (int S = 1; S <= all; S++)
      H[i][S] = H[i][(S - 1) & S] * H[i][-S & S] % MOD;
  }
  int ans = 0;
  for (int S = 1; S < 1 << n; S++) {
    g[0] = 1;
    for (int tS = (S - 1) & S; tS; tS = (tS - 1) & S) {
      int mask = S ^ tS;
      int x = __builtin_ffs(mask) - 1;
      g[mask] = g[mask ^ (1 << x)] * H[x][all ^ S] % MOD;
      f[S] = (f[S] - f[tS] * g[mask]) % MOD;
    }
    int x = __builtin_ffs(S) - 1;
    g[S] = g[S ^ (1 << x)] * H[x][all ^ S] % MOD;
    f[S] = (f[S] + g[S]) % MOD;
    ans = (ans + f[S] * __builtin_popcount(S)) % MOD;
  }
  cout << (ans + MOD) % MOD << "\n";
}

```