#include <bits/stdc++.h>
using namespace std;
const int N = 15, M = 1 << 15;
const long long mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
long long dp[M], val[N][M], sum[N][M], a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      val[i][j] = a[i] * qpow(a[i] + a[j], mod - 2) % mod;
  int mx = 1 << n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < mx; j++) {
      sum[i][j] = 1;
      for (int k = 0; k < n; k++)
        if ((1 << k) & j) sum[i][j] = sum[i][j] * val[i][k] % mod;
    }
  }
  for (int i = 1; i < mx; i++) {
    for (int j = 0; j < n; j++)
      if (i & (1 << j)) dp[i]++;
    for (int j = i & (i - 1); j; j = (j - 1) & i) {
      long long del = dp[j];
      for (int k = 0; k < n; k++)
        if (((1 << k) & i) && !((1 << k) & j)) del = del * sum[k][j] % mod;
      dp[i] = (dp[i] - del + mod) % mod;
    }
  }
  printf("%lld\n", dp[mx - 1]);
  return 0;
}

### 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 = 15, M = 1 << 15;
const long long mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
long long dp[M], val[N][M], sum[N][M], a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      val[i][j] = a[i] * qpow(a[i] + a[j], mod - 2) % mod;
  int mx = 1 << n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < mx; j++) {
      sum[i][j] = 1;
      for (int k = 0; k < n; k++)
        if ((1 << k) & j) sum[i][j] = sum[i][j] * val[i][k] % mod;
    }
  }
  for (int i = 1; i < mx; i++) {
    for (int j = 0; j < n; j++)
      if (i & (1 << j)) dp[i]++;
    for (int j = i & (i - 1); j; j = (j - 1) & i) {
      long long del = dp[j];
      for (int k = 0; k < n; k++)
        if (((1 << k) & i) && !((1 << k) & j)) del = del * sum[k][j] % mod;
      dp[i] = (dp[i] - del + mod) % mod;
    }
  }
  printf("%lld\n", dp[mx - 1]);
  return 0;
}

```