#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int n;
long long a[20], pr[20][20];
long long dp[(1 << 14) + 10];
long long pot(long long u, long long v) {
  u %= mod;
  long long z = 1;
  while (v) {
    if (v & 1) {
      z = (z * u) % mod;
    }
    u = (u * u) % mod;
    v /= 2;
  }
  return z;
}
long long inv(long long u) { return pot(u, mod - 2); }
long long f(int i, int j) {
  long long ans = 1;
  for (int x = 0; x < n; ++x) {
    if ((1 << x) & i && !((1 << x) & j)) {
      long long cur = 1;
      for (int k = 0; k < n; ++k) {
        if ((1 << k) & i) {
          continue;
        }
        cur = (cur * pr[x][k]) % mod;
      }
      ans = (ans * (1 - cur + mod)) % mod;
    }
  }
  if (__builtin_popcount(i ^ j) % 2 == 0) {
    ans = (mod - ans) % mod;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      pr[i][j] = (a[i] * inv(a[i] + a[j])) % mod;
    }
  }
  int l = (1 << n) - 1;
  dp[l] = 1;
  for (int i = 1; i < (1 << n); ++i) {
    int j = i;
    do {
      j = (j - 1) & i;
      dp[l ^ i] += dp[l ^ j] * f(i, j);
      dp[l ^ i] %= mod;
    } while (j > 0);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += dp[(1 << i)];
    ans %= mod;
  }
  printf("%lld\n", ans);
}

### 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;
int n;
long long a[20], pr[20][20];
long long dp[(1 << 14) + 10];
long long pot(long long u, long long v) {
  u %= mod;
  long long z = 1;
  while (v) {
    if (v & 1) {
      z = (z * u) % mod;
    }
    u = (u * u) % mod;
    v /= 2;
  }
  return z;
}
long long inv(long long u) { return pot(u, mod - 2); }
long long f(int i, int j) {
  long long ans = 1;
  for (int x = 0; x < n; ++x) {
    if ((1 << x) & i && !((1 << x) & j)) {
      long long cur = 1;
      for (int k = 0; k < n; ++k) {
        if ((1 << k) & i) {
          continue;
        }
        cur = (cur * pr[x][k]) % mod;
      }
      ans = (ans * (1 - cur + mod)) % mod;
    }
  }
  if (__builtin_popcount(i ^ j) % 2 == 0) {
    ans = (mod - ans) % mod;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      pr[i][j] = (a[i] * inv(a[i] + a[j])) % mod;
    }
  }
  int l = (1 << n) - 1;
  dp[l] = 1;
  for (int i = 1; i < (1 << n); ++i) {
    int j = i;
    do {
      j = (j - 1) & i;
      dp[l ^ i] += dp[l ^ j] * f(i, j);
      dp[l ^ i] %= mod;
    } while (j > 0);
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += dp[(1 << i)];
    ans %= mod;
  }
  printf("%lld\n", ans);
}

```