#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long modpow(long long a, long long x) {
  long long tmp = x;
  long long val = a;
  long long ret = 1LL;
  while (tmp > 0) {
    if (tmp % 2 == 1) ret = ret * val % mod;
    val = val * val % mod;
    tmp /= 2;
  }
  return ret;
}
long long modinv(long long a) { return modpow(a, mod - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> G(1 << n, vector<long long>(n, 1));
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if (((i >> j) & 1) == 0) continue;
      for (int k = 0; k < n; k++) {
        G[i][k] = G[i][k] * a[j] % mod * modinv(a[j] + a[k]) % mod;
      }
    }
  }
  long long ans = 0LL;
  vector<long long> P(1 << n);
  for (int i = 1; i < (1 << n); i++) {
    long long cnt = __builtin_popcount(i);
    long long P_tmp = 1LL;
    for (int j = i; j > 0; j = ((j - 1) & i)) {
      if (j == i) continue;
      long long G_tmp0 = 1LL;
      for (int k = 0; k < n; k++) {
        if (((j >> k) & 1) == 1) continue;
        if (((i >> k) & 1) == 0) continue;
        G_tmp0 = (G_tmp0 * G[j][k]) % mod;
      }
      P_tmp = (P_tmp - P[j] * G_tmp0 % mod + mod) % mod;
    }
    long long G_tmp1 = 1LL;
    for (int k = 0; k < n; k++) {
      if (((i >> k) & 1) == 1) continue;
      G_tmp1 = (G_tmp1 * G[i][k]) % mod;
    }
    ans = (ans + cnt * P_tmp % mod * G_tmp1) % mod;
    P[i] = P_tmp;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate a CPP solution to 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 long long mod = 1e9 + 7;
long long modpow(long long a, long long x) {
  long long tmp = x;
  long long val = a;
  long long ret = 1LL;
  while (tmp > 0) {
    if (tmp % 2 == 1) ret = ret * val % mod;
    val = val * val % mod;
    tmp /= 2;
  }
  return ret;
}
long long modinv(long long a) { return modpow(a, mod - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> G(1 << n, vector<long long>(n, 1));
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if (((i >> j) & 1) == 0) continue;
      for (int k = 0; k < n; k++) {
        G[i][k] = G[i][k] * a[j] % mod * modinv(a[j] + a[k]) % mod;
      }
    }
  }
  long long ans = 0LL;
  vector<long long> P(1 << n);
  for (int i = 1; i < (1 << n); i++) {
    long long cnt = __builtin_popcount(i);
    long long P_tmp = 1LL;
    for (int j = i; j > 0; j = ((j - 1) & i)) {
      if (j == i) continue;
      long long G_tmp0 = 1LL;
      for (int k = 0; k < n; k++) {
        if (((j >> k) & 1) == 1) continue;
        if (((i >> k) & 1) == 0) continue;
        G_tmp0 = (G_tmp0 * G[j][k]) % mod;
      }
      P_tmp = (P_tmp - P[j] * G_tmp0 % mod + mod) % mod;
    }
    long long G_tmp1 = 1LL;
    for (int k = 0; k < n; k++) {
      if (((i >> k) & 1) == 1) continue;
      G_tmp1 = (G_tmp1 * G[i][k]) % mod;
    }
    ans = (ans + cnt * P_tmp % mod * G_tmp1) % mod;
    P[i] = P_tmp;
  }
  cout << ans << "\n";
  return 0;
}

```