#include <bits/stdc++.h>
using namespace std;
const int N = 15;
const long long mod = 1e9 + 7;
int n;
int a[N];
int g[1 << N | 1], f[1 << N | 1];
int inv[N][N];
inline int Pop(int x) {
  int c = 0;
  while (x) ++c, x -= x & -x;
  return c;
}
inline int power(int a, int p) {
  int ans = 1;
  for (; p; p >>= 1)
    (p & 1) && (ans = 1ll * ans * a % mod), a = 1ll * a * a % mod;
  return ans;
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (register int i = 1; i <= n; ++i)
    for (register int j = 1; j <= n; ++j)
      inv[i][j] = power(a[i] + a[j], mod - 2);
  for (register int sta = 1; sta < 1 << n; ++sta) {
    g[sta] = 1;
    for (register int i = 1; i <= n; ++i) {
      if (!(sta & (1 << (i - 1)))) continue;
      for (register int j = 1; j <= n; ++j) {
        if ((sta & (1 << (j - 1)))) continue;
        g[sta] = 1ll * g[sta] * a[i] % mod * inv[i][j] % mod;
      }
    }
  }
  for (register int S = 1; S < 1 << n; ++S) {
    f[S] = mod - g[S];
    for (register int T = S; T; T = (T - 1) & S) {
      int iv = 1;
      for (register int i = 1; i <= n; ++i) {
        if (!(S & (1 << i - 1) && !(T & (1 << i - 1)))) continue;
        for (register int j = 1; j <= n; ++j) {
          if (!(T & (1 << j - 1))) continue;
          iv = 1ll * a[i] * inv[i][j] % mod * iv % mod;
        }
      }
      f[S] = (f[S] + 1ll * f[T] * g[S xor T] % mod * power(iv, mod - 2)) % mod;
    }
    f[S] = mod - f[S];
  }
  int ans = 0;
  for (register int S = 1; S < 1 << n; ++S)
    ans = (ans + 1ll * Pop(S) * f[S]) % mod;
  printf("%d\n", ans);
  return 0;
}

### 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 = 15;
const long long mod = 1e9 + 7;
int n;
int a[N];
int g[1 << N | 1], f[1 << N | 1];
int inv[N][N];
inline int Pop(int x) {
  int c = 0;
  while (x) ++c, x -= x & -x;
  return c;
}
inline int power(int a, int p) {
  int ans = 1;
  for (; p; p >>= 1)
    (p & 1) && (ans = 1ll * ans * a % mod), a = 1ll * a * a % mod;
  return ans;
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (register int i = 1; i <= n; ++i)
    for (register int j = 1; j <= n; ++j)
      inv[i][j] = power(a[i] + a[j], mod - 2);
  for (register int sta = 1; sta < 1 << n; ++sta) {
    g[sta] = 1;
    for (register int i = 1; i <= n; ++i) {
      if (!(sta & (1 << (i - 1)))) continue;
      for (register int j = 1; j <= n; ++j) {
        if ((sta & (1 << (j - 1)))) continue;
        g[sta] = 1ll * g[sta] * a[i] % mod * inv[i][j] % mod;
      }
    }
  }
  for (register int S = 1; S < 1 << n; ++S) {
    f[S] = mod - g[S];
    for (register int T = S; T; T = (T - 1) & S) {
      int iv = 1;
      for (register int i = 1; i <= n; ++i) {
        if (!(S & (1 << i - 1) && !(T & (1 << i - 1)))) continue;
        for (register int j = 1; j <= n; ++j) {
          if (!(T & (1 << j - 1))) continue;
          iv = 1ll * a[i] * inv[i][j] % mod * iv % mod;
        }
      }
      f[S] = (f[S] + 1ll * f[T] * g[S xor T] % mod * power(iv, mod - 2)) % mod;
    }
    f[S] = mod - f[S];
  }
  int ans = 0;
  for (register int S = 1; S < 1 << n; ++S)
    ans = (ans + 1ll * Pop(S) * f[S]) % mod;
  printf("%d\n", ans);
  return 0;
}

```