#include <bits/stdc++.h>
using namespace std;
long long exp(long long x, long long y) {
  long long MOD = 1000000007;
  if (y == 1) {
    return x;
  }
  long long z = exp(x, y / 2);
  z = (z * z) % MOD;
  if (y % 2 == 1) {
    z = (z * x) % MOD;
  }
  return z;
}
long long inv(long long x) { return exp(x, 1000000005); }
int main() {
  long long n;
  cin >> n;
  long long MOD = 1000000007;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long b[n][n];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      b[i][j] = (a[j] * inv(a[i] + a[j])) % MOD;
    }
  }
  long long m = 1 << n;
  long long c[n][m];
  for (long long i = 0; i < n; i++) {
    c[i][0] = 1;
    for (long long j = 0; j < n; j++) {
      for (long long k = 0; k < (1 << j); k++) {
        c[i][k + (1 << j)] = (c[i][k] * b[i][j]) % MOD;
      }
    }
  }
  long long d[m];
  d[0] = 0;
  for (long long i = 1; i < m; i++) {
    bool e[n];
    long long t = 0;
    for (long long j = 0; j < n; j++) {
      if ((i & (1 << j)) == 0) {
        e[j] = false;
      } else {
        e[j] = true;
        t++;
      }
    }
    long long g[t];
    long long temp = 0;
    for (long long j = 0; j < n; j++) {
      if (e[j]) {
        g[temp] = j;
        temp++;
      }
    }
    long long f[1 << t];
    f[0] = 0;
    for (long long j = 0; j < t; j++) {
      for (long long k = 0; k < (1 << j); k++) {
        f[k + (1 << j)] = f[k] + (1 << g[j]);
      }
    }
    long long v = 0;
    for (long long j = 0; j < (1 << t) - 1; j++) {
      long long w = d[f[j]];
      for (long long k = 0; k < n; k++) {
        if (e[k] && ((f[j] & (1 << k)) == 0)) {
          w = (w * c[k][f[j]]) % MOD;
        }
      }
      v = (v + w) % MOD;
    }
    d[i] = (MOD + 1 - v) % MOD;
  }
  long long ans = 0;
  for (long long i = 0; i < m; i++) {
    long long w = d[i];
    long long t = 0;
    for (long long k = 0; k < n; k++) {
      if ((i & (1 << k)) == 0) {
        w = (w * c[k][i]) % MOD;
      } else {
        t++;
      }
    }
    ans = (ans + (w * t)) % MOD;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long exp(long long x, long long y) {
  long long MOD = 1000000007;
  if (y == 1) {
    return x;
  }
  long long z = exp(x, y / 2);
  z = (z * z) % MOD;
  if (y % 2 == 1) {
    z = (z * x) % MOD;
  }
  return z;
}
long long inv(long long x) { return exp(x, 1000000005); }
int main() {
  long long n;
  cin >> n;
  long long MOD = 1000000007;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long b[n][n];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      b[i][j] = (a[j] * inv(a[i] + a[j])) % MOD;
    }
  }
  long long m = 1 << n;
  long long c[n][m];
  for (long long i = 0; i < n; i++) {
    c[i][0] = 1;
    for (long long j = 0; j < n; j++) {
      for (long long k = 0; k < (1 << j); k++) {
        c[i][k + (1 << j)] = (c[i][k] * b[i][j]) % MOD;
      }
    }
  }
  long long d[m];
  d[0] = 0;
  for (long long i = 1; i < m; i++) {
    bool e[n];
    long long t = 0;
    for (long long j = 0; j < n; j++) {
      if ((i & (1 << j)) == 0) {
        e[j] = false;
      } else {
        e[j] = true;
        t++;
      }
    }
    long long g[t];
    long long temp = 0;
    for (long long j = 0; j < n; j++) {
      if (e[j]) {
        g[temp] = j;
        temp++;
      }
    }
    long long f[1 << t];
    f[0] = 0;
    for (long long j = 0; j < t; j++) {
      for (long long k = 0; k < (1 << j); k++) {
        f[k + (1 << j)] = f[k] + (1 << g[j]);
      }
    }
    long long v = 0;
    for (long long j = 0; j < (1 << t) - 1; j++) {
      long long w = d[f[j]];
      for (long long k = 0; k < n; k++) {
        if (e[k] && ((f[j] & (1 << k)) == 0)) {
          w = (w * c[k][f[j]]) % MOD;
        }
      }
      v = (v + w) % MOD;
    }
    d[i] = (MOD + 1 - v) % MOD;
  }
  long long ans = 0;
  for (long long i = 0; i < m; i++) {
    long long w = d[i];
    long long t = 0;
    for (long long k = 0; k < n; k++) {
      if ((i & (1 << k)) == 0) {
        w = (w * c[k][i]) % MOD;
      } else {
        t++;
      }
    }
    ans = (ans + (w * t)) % MOD;
  }
  cout << ans << '\n';
  return 0;
}

```