#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007ll;
const int N = 14;
long long power(long long a, long long x) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans = ans * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return ans;
}
long long inv(long long x) { return power(x, mod - 2); }
long long h[N + 5][1 << N];
long long p[N + 5][N + 5];
long long a[N + 5], f[1 << N];
long long g[N + 5];
int cal(int x) {
  int ans = 0;
  while (x) {
    ans += x & 1;
    x >>= 1;
  }
  return ans;
}
long long temp[1 << N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) p[i][j] = a[i] * inv(a[i] + a[j]) % mod;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < (1 << n); j++) {
      if (j & (1 << (i - 1))) continue;
      h[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if (j & (1 << (k - 1))) h[i][j] = h[i][j] * p[i][k] % mod;
    }
  long long ans = 0;
  for (int i = 1; i < 1 << n; i++) {
    g[i] = 1;
    for (int j = i & (i - 1); j; j = (j - 1) & i) {
      for (int k = 1; k <= n; k++)
        if ((i & (1 << (k - 1))) && !(j & (1 << (k - 1)))) {
          g[j] = g[j ^ (1 << (k - 1))] * h[k][((1 << n) - 1) ^ i] % mod;
          break;
        }
      f[i] = (f[i] + (mod - temp[j]) * g[j] % mod) % mod;
    }
    for (int k = 1; k <= n; k++)
      if ((i & (1 << (k - 1)))) {
        g[0] = g[(1 << (k - 1))] * h[k][((1 << n) - 1) ^ i] % mod;
        break;
      }
    f[i] = (f[i] + g[0]) % mod;
    temp[i] = f[i];
    ans = (ans + f[i] * cal(i) % mod) % mod;
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Your task is to create 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 long long mod = 1000000007ll;
const int N = 14;
long long power(long long a, long long x) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans = ans * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return ans;
}
long long inv(long long x) { return power(x, mod - 2); }
long long h[N + 5][1 << N];
long long p[N + 5][N + 5];
long long a[N + 5], f[1 << N];
long long g[N + 5];
int cal(int x) {
  int ans = 0;
  while (x) {
    ans += x & 1;
    x >>= 1;
  }
  return ans;
}
long long temp[1 << N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) p[i][j] = a[i] * inv(a[i] + a[j]) % mod;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < (1 << n); j++) {
      if (j & (1 << (i - 1))) continue;
      h[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if (j & (1 << (k - 1))) h[i][j] = h[i][j] * p[i][k] % mod;
    }
  long long ans = 0;
  for (int i = 1; i < 1 << n; i++) {
    g[i] = 1;
    for (int j = i & (i - 1); j; j = (j - 1) & i) {
      for (int k = 1; k <= n; k++)
        if ((i & (1 << (k - 1))) && !(j & (1 << (k - 1)))) {
          g[j] = g[j ^ (1 << (k - 1))] * h[k][((1 << n) - 1) ^ i] % mod;
          break;
        }
      f[i] = (f[i] + (mod - temp[j]) * g[j] % mod) % mod;
    }
    for (int k = 1; k <= n; k++)
      if ((i & (1 << (k - 1)))) {
        g[0] = g[(1 << (k - 1))] * h[k][((1 << n) - 1) ^ i] % mod;
        break;
      }
    f[i] = (f[i] + g[0]) % mod;
    temp[i] = f[i];
    ans = (ans + f[i] * cal(i) % mod) % mod;
  }
  printf("%lld", ans);
  return 0;
}

```