#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
const int maxv = 1e6 + 10;
const int maxn = 2e5 + 10;
const int inf32 = 1e9 + 10;
const long long inf64 = 1e18 + 10;
const int mod = 1e9 + 7;
long long mul(long long first, long long second) {
  return first * second % mod;
}
long long add(long long first, long long second) {
  return (first + second) % mod;
}
long long sub(long long first, long long second) {
  return ((first - second) % mod + mod) % mod;
}
long long qpow(long long first, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = mul(ans, first);
    first = mul(first, first);
    n /= 2;
  }
  return ans;
}
long long inverse(long long first) { return qpow(first, mod - 2); }
int N, a[14];
long long inv[maxv * 2], g[14][1 << 14], dp[1 << 14];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> N;
  for (int i = 0; i < N; ++i) cin >> a[i];
  for (int i = 1; i < maxv * 2; ++i) inv[i] = inverse(i);
  for (int i = 0; i < N; ++i) {
    for (int mask = 0; mask < 1 << N; ++mask) {
      if (mask >> i & 1) continue;
      g[i][mask] = 1;
      for (int j = 0; j < N; ++j) {
        if (mask >> j & 1) {
          g[i][mask] = mul(g[i][mask], mul(a[i], inv[a[i] + a[j]]));
        }
      }
    }
  }
  for (int mask = 1; mask < 1 << N; ++mask) {
    dp[mask] = 1;
    for (int submask = mask - 1; submask > 0; submask = (submask - 1) & mask) {
      long long v = 1;
      for (int i = 0; i < N; ++i) {
        if (submask >> i & 1) {
          v = mul(v, g[i][mask ^ submask]);
        }
      }
      dp[mask] = sub(dp[mask], mul(dp[submask], v));
    }
  }
  long long ans = 0;
  for (int mask = 1; mask < 1 << N; ++mask) {
    long long v = 1;
    for (int i = 0; i < N; ++i) {
      if (mask >> i & 1) {
        v = mul(v, g[i][((1 << N) - 1) ^ mask]);
      }
    }
    ans = add(ans, mul(mul(dp[mask], v), __builtin_popcount(mask)));
  }
  cout << ans << endl;
}

### Prompt

Please create a solution in Cpp 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>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
const int maxv = 1e6 + 10;
const int maxn = 2e5 + 10;
const int inf32 = 1e9 + 10;
const long long inf64 = 1e18 + 10;
const int mod = 1e9 + 7;
long long mul(long long first, long long second) {
  return first * second % mod;
}
long long add(long long first, long long second) {
  return (first + second) % mod;
}
long long sub(long long first, long long second) {
  return ((first - second) % mod + mod) % mod;
}
long long qpow(long long first, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = mul(ans, first);
    first = mul(first, first);
    n /= 2;
  }
  return ans;
}
long long inverse(long long first) { return qpow(first, mod - 2); }
int N, a[14];
long long inv[maxv * 2], g[14][1 << 14], dp[1 << 14];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> N;
  for (int i = 0; i < N; ++i) cin >> a[i];
  for (int i = 1; i < maxv * 2; ++i) inv[i] = inverse(i);
  for (int i = 0; i < N; ++i) {
    for (int mask = 0; mask < 1 << N; ++mask) {
      if (mask >> i & 1) continue;
      g[i][mask] = 1;
      for (int j = 0; j < N; ++j) {
        if (mask >> j & 1) {
          g[i][mask] = mul(g[i][mask], mul(a[i], inv[a[i] + a[j]]));
        }
      }
    }
  }
  for (int mask = 1; mask < 1 << N; ++mask) {
    dp[mask] = 1;
    for (int submask = mask - 1; submask > 0; submask = (submask - 1) & mask) {
      long long v = 1;
      for (int i = 0; i < N; ++i) {
        if (submask >> i & 1) {
          v = mul(v, g[i][mask ^ submask]);
        }
      }
      dp[mask] = sub(dp[mask], mul(dp[submask], v));
    }
  }
  long long ans = 0;
  for (int mask = 1; mask < 1 << N; ++mask) {
    long long v = 1;
    for (int i = 0; i < N; ++i) {
      if (mask >> i & 1) {
        v = mul(v, g[i][((1 << N) - 1) ^ mask]);
      }
    }
    ans = add(ans, mul(mul(dp[mask], v), __builtin_popcount(mask)));
  }
  cout << ans << endl;
}

```