#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 15;
const int M = 1 << N;
int a[N], dp[M], p[N][M];
int add(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return 1ll * a * b % mod; }
int qpow(int a, int n) {
  int re = 1;
  while (n > 0) {
    if (n & 1) {
      re = mul(re, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return re;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      p[j][i] = 1;
    }
    for (int j = 0; j < n; j++) {
      if (i >> j & 1) {
        for (int k = 0; k < n; k++) {
          if (i >> k & 1) continue;
          p[k][i] =
              mul(p[k][i ^ 1 << j], mul(a[k], qpow(add(a[k], a[j]), mod - 2)));
        }
        break;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < (1 << n); i++) {
    dp[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      int tmp = 1;
      for (int k = 0; k < n; k++) {
        if (j >> k & 1) {
          tmp = mul(tmp, p[k][i ^ j]);
        }
      }
      dp[i] = add(dp[i], mod - mul(dp[j], tmp));
    }
    int tmp = 1;
    for (int j = 0; j < n; j++) {
      if (i >> j & 1) {
        tmp = mul(tmp, p[j][(1 << n) - 1 ^ i]);
      }
    }
    ans = add(ans, mul(mul(dp[i], tmp), __builtin_popcount(i)));
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

<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 mod = 1e9 + 7;
const int N = 15;
const int M = 1 << N;
int a[N], dp[M], p[N][M];
int add(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return 1ll * a * b % mod; }
int qpow(int a, int n) {
  int re = 1;
  while (n > 0) {
    if (n & 1) {
      re = mul(re, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return re;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      p[j][i] = 1;
    }
    for (int j = 0; j < n; j++) {
      if (i >> j & 1) {
        for (int k = 0; k < n; k++) {
          if (i >> k & 1) continue;
          p[k][i] =
              mul(p[k][i ^ 1 << j], mul(a[k], qpow(add(a[k], a[j]), mod - 2)));
        }
        break;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < (1 << n); i++) {
    dp[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      int tmp = 1;
      for (int k = 0; k < n; k++) {
        if (j >> k & 1) {
          tmp = mul(tmp, p[k][i ^ j]);
        }
      }
      dp[i] = add(dp[i], mod - mul(dp[j], tmp));
    }
    int tmp = 1;
    for (int j = 0; j < n; j++) {
      if (i >> j & 1) {
        tmp = mul(tmp, p[j][(1 << n) - 1 ^ i]);
      }
    }
    ans = add(ans, mul(mul(dp[i], tmp), __builtin_popcount(i)));
  }
  printf("%d", ans);
  return 0;
}

```