#include <bits/stdc++.h>
using namespace std;
long long fastxp(long long n, long long e) {
  long long ans = 1, pwr = n;
  while (e) {
    if (e % 2) ans = ans * pwr % 1000000007;
    e /= 2;
    pwr = pwr * pwr % 1000000007;
  }
  return ans;
}
long long n;
long long a[20], win[20][20], dp[1 << 14];
long long win_all(long long m1, long long m2) {
  long long ret = 1;
  for (long long i = 0; i < n; i++)
    if ((1 << i) & m1)
      for (long long j = 0; j < n; j++)
        if ((1 << j) & m2) {
          ret = ret * win[i][j] % 1000000007;
        }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++)
      win[i][j] = a[i] * fastxp(a[i] + a[j], 1000000007 - 2) % 1000000007;
  long long ans = 0;
  for (long long mask = 0; mask < (1 << n); mask++) {
    dp[mask] = 1;
    for (long long i = mask & (mask - 1); i > 0; i = (i - 1) & mask) {
      dp[mask] = (dp[mask] - win_all(i, mask ^ i) * dp[i]) % 1000000007;
    }
    dp[mask] = (dp[mask] + 1000000007) % 1000000007;
    ans = (__builtin_popcount(mask) * dp[mask] % 1000000007 *
               win_all(mask, ((1 << n) - 1) ^ mask) +
           ans) %
          1000000007;
  }
  cout << ans << endl;
}

### 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;
long long fastxp(long long n, long long e) {
  long long ans = 1, pwr = n;
  while (e) {
    if (e % 2) ans = ans * pwr % 1000000007;
    e /= 2;
    pwr = pwr * pwr % 1000000007;
  }
  return ans;
}
long long n;
long long a[20], win[20][20], dp[1 << 14];
long long win_all(long long m1, long long m2) {
  long long ret = 1;
  for (long long i = 0; i < n; i++)
    if ((1 << i) & m1)
      for (long long j = 0; j < n; j++)
        if ((1 << j) & m2) {
          ret = ret * win[i][j] % 1000000007;
        }
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++)
      win[i][j] = a[i] * fastxp(a[i] + a[j], 1000000007 - 2) % 1000000007;
  long long ans = 0;
  for (long long mask = 0; mask < (1 << n); mask++) {
    dp[mask] = 1;
    for (long long i = mask & (mask - 1); i > 0; i = (i - 1) & mask) {
      dp[mask] = (dp[mask] - win_all(i, mask ^ i) * dp[i]) % 1000000007;
    }
    dp[mask] = (dp[mask] + 1000000007) % 1000000007;
    ans = (__builtin_popcount(mask) * dp[mask] % 1000000007 *
               win_all(mask, ((1 << n) - 1) ^ mask) +
           ans) %
          1000000007;
  }
  cout << ans << endl;
}

```