#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long n, a[15], win[15][15];
long long mult(long long x, long long k) {
  if (k == 0) return 1;
  long long tmp = mult(x, k / 2);
  if (k % 2 == 0) return tmp * tmp % MOD;
  return tmp * tmp % MOD * x % MOD;
}
long long getBit(long long x, long long k) { return (x >> k) & 1; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  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++)
      if (i != j) win[i][j] = a[i] * mult(a[i] + a[j], MOD - 2) % MOD;
  vector<long long> f(1 << n, 1);
  for (long long mask = 1; mask < (1 << n); mask++)
    for (long long submask = (mask - 1) & mask; submask;
         submask = (submask - 1) & mask) {
      long long tmp = f[submask];
      for (long long i = 0; i < n; i++)
        if (getBit(submask, i))
          for (long long j = 0; j < n; j++)
            if (getBit(mask, j) && !getBit(submask, j))
              tmp = tmp * win[i][j] % MOD;
      f[mask] = (f[mask] - tmp + MOD) % MOD;
    }
  long long rs = 0;
  for (long long mask = 1; mask < (1 << n); mask++) {
    long long tmp = f[mask];
    for (long long i = 0; i < n; i++)
      if (getBit(mask, i))
        for (long long j = 0; j < n; j++)
          if (!getBit(mask, j)) tmp = tmp * win[i][j] % MOD;
    rs = (rs + __builtin_popcount(mask) * tmp) % MOD;
  }
  cout << rs;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

<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 = 1e9 + 7;
long long n, a[15], win[15][15];
long long mult(long long x, long long k) {
  if (k == 0) return 1;
  long long tmp = mult(x, k / 2);
  if (k % 2 == 0) return tmp * tmp % MOD;
  return tmp * tmp % MOD * x % MOD;
}
long long getBit(long long x, long long k) { return (x >> k) & 1; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  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++)
      if (i != j) win[i][j] = a[i] * mult(a[i] + a[j], MOD - 2) % MOD;
  vector<long long> f(1 << n, 1);
  for (long long mask = 1; mask < (1 << n); mask++)
    for (long long submask = (mask - 1) & mask; submask;
         submask = (submask - 1) & mask) {
      long long tmp = f[submask];
      for (long long i = 0; i < n; i++)
        if (getBit(submask, i))
          for (long long j = 0; j < n; j++)
            if (getBit(mask, j) && !getBit(submask, j))
              tmp = tmp * win[i][j] % MOD;
      f[mask] = (f[mask] - tmp + MOD) % MOD;
    }
  long long rs = 0;
  for (long long mask = 1; mask < (1 << n); mask++) {
    long long tmp = f[mask];
    for (long long i = 0; i < n; i++)
      if (getBit(mask, i))
        for (long long j = 0; j < n; j++)
          if (!getBit(mask, j)) tmp = tmp * win[i][j] % MOD;
    rs = (rs + __builtin_popcount(mask) * tmp) % MOD;
  }
  cout << rs;
}

```