#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int MOD = 1e9 + 7;
long long a[20], dp[20010], prob[20][20], n;
long long pw(long long a, long long n) {
  long long base = a, res = 1;
  while (n) {
    if (n & 1) res = res * base % MOD;
    base = base * base % MOD;
    n >>= 1;
  }
  return res;
}
long long inv(long long a) { return pw(a, MOD - 2); }
long long get_dp(int x) {
  if (dp[x] >= 0) return dp[x];
  vector<int> b;
  for (int i = 0; i < n; i++) {
    if ((1 << i) & x) b.push_back(i);
  }
  long long res = 1;
  int m = b.size();
  if (m == 1) {
    dp[x] = 1;
    return dp[x];
  }
  for (int i = 1; i < (1 << m) - 1; i++) {
    int y = 0;
    for (int j = 0; j < m; j++) {
      if (i & (1 << j)) y |= (1 << b[j]);
    }
    long long t = get_dp(y);
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < m; k++) {
        if (((1 << j) & i) == 0 && ((1 << k) & i) != 0)
          t = t * prob[b[j]][b[k]] % MOD;
      }
    }
    res = res - t % MOD;
    res = (res + MOD) % MOD;
  }
  dp[x] = res;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j)
        prob[i][j] = 1;
      else {
        prob[i][j] = a[i] * inv(a[i] + a[j]) % MOD;
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0] = 1;
  for (int i = 1; i < (1 << n); i++) {
    get_dp(i);
  }
  long long ans = n;
  for (int i = 1; i < (1 << n) - 1; i++) {
    long long cnt = 0;
    for (int j = 0; j < n; j++) {
      if ((i & (1 << j)) == 0) {
        cnt++;
      }
    }
    long long tmp = dp[i];
    for (int k = 0; k < n; k++) {
      for (int l = 0; l < n; l++)
        if ((i & (1 << l)) == 0 && (i & (1 << k))) {
          tmp = tmp * prob[k][l] % MOD;
        }
    }
    ans -= tmp * cnt % MOD;
    ans %= MOD;
    ans = (ans + MOD) % MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 maxn = 1e5 + 10;
const int MOD = 1e9 + 7;
long long a[20], dp[20010], prob[20][20], n;
long long pw(long long a, long long n) {
  long long base = a, res = 1;
  while (n) {
    if (n & 1) res = res * base % MOD;
    base = base * base % MOD;
    n >>= 1;
  }
  return res;
}
long long inv(long long a) { return pw(a, MOD - 2); }
long long get_dp(int x) {
  if (dp[x] >= 0) return dp[x];
  vector<int> b;
  for (int i = 0; i < n; i++) {
    if ((1 << i) & x) b.push_back(i);
  }
  long long res = 1;
  int m = b.size();
  if (m == 1) {
    dp[x] = 1;
    return dp[x];
  }
  for (int i = 1; i < (1 << m) - 1; i++) {
    int y = 0;
    for (int j = 0; j < m; j++) {
      if (i & (1 << j)) y |= (1 << b[j]);
    }
    long long t = get_dp(y);
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < m; k++) {
        if (((1 << j) & i) == 0 && ((1 << k) & i) != 0)
          t = t * prob[b[j]][b[k]] % MOD;
      }
    }
    res = res - t % MOD;
    res = (res + MOD) % MOD;
  }
  dp[x] = res;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j)
        prob[i][j] = 1;
      else {
        prob[i][j] = a[i] * inv(a[i] + a[j]) % MOD;
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0] = 1;
  for (int i = 1; i < (1 << n); i++) {
    get_dp(i);
  }
  long long ans = n;
  for (int i = 1; i < (1 << n) - 1; i++) {
    long long cnt = 0;
    for (int j = 0; j < n; j++) {
      if ((i & (1 << j)) == 0) {
        cnt++;
      }
    }
    long long tmp = dp[i];
    for (int k = 0; k < n; k++) {
      for (int l = 0; l < n; l++)
        if ((i & (1 << l)) == 0 && (i & (1 << k))) {
          tmp = tmp * prob[k][l] % MOD;
        }
    }
    ans -= tmp * cnt % MOD;
    ans %= MOD;
    ans = (ans + MOD) % MOD;
  }
  cout << ans << endl;
  return 0;
}

```