#include <bits/stdc++.h>
using namespace std;
template <class T>
using v2d = vector<vector<T> >;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 14;
const long long mod = 1e9 + 7;
int n;
long long a[14], f[1 << 14], w[14][14], h[14][1 << 14];
long long power(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) {
      r = r * a % mod;
    }
    b /= 2;
    a = a * a % mod;
  }
  return r;
}
void solve() {
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(n); ++j) {
      w[i][j] = a[i] * power(a[i] + a[j], mod - 2) % mod;
    }
    for (int mask = 0; mask < (int)(1 << n); ++mask) {
      h[i][mask] = 1;
      for (int j = 0; j < (int)(n); ++j) {
        if (mask >> j & 1) {
          h[i][mask] = h[i][mask] * w[i][j] % mod;
        }
      }
    }
  }
  int m = (1 << n) - 1;
  for (int mask = 1; mask <= (int)(m); ++mask) {
    f[mask] = 1;
    int submask = (mask - 1) & mask;
    while (submask) {
      long long p = 1;
      for (int i = 0; i < (int)(n); ++i) {
        if (submask >> i & 1) {
          p = p * h[i][mask ^ submask] % mod;
        }
      }
      f[mask] = (f[mask] - p * f[submask] + mod * mod) % mod;
      submask = (submask - 1) & mask;
    }
  }
  long long ans = 0;
  for (int mask = 1; mask <= (int)(m); ++mask) {
    int cnt = __builtin_popcount(mask);
    long long p = 1;
    for (int i = 0; i < (int)(n); ++i) {
      if (mask >> i & 1) {
        p = p * h[i][m ^ mask] % mod;
      }
    }
    ans = (ans + cnt * f[mask] % mod * p) % mod;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### 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;
template <class T>
using v2d = vector<vector<T> >;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 14;
const long long mod = 1e9 + 7;
int n;
long long a[14], f[1 << 14], w[14][14], h[14][1 << 14];
long long power(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) {
      r = r * a % mod;
    }
    b /= 2;
    a = a * a % mod;
  }
  return r;
}
void solve() {
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = 0; j < (int)(n); ++j) {
      w[i][j] = a[i] * power(a[i] + a[j], mod - 2) % mod;
    }
    for (int mask = 0; mask < (int)(1 << n); ++mask) {
      h[i][mask] = 1;
      for (int j = 0; j < (int)(n); ++j) {
        if (mask >> j & 1) {
          h[i][mask] = h[i][mask] * w[i][j] % mod;
        }
      }
    }
  }
  int m = (1 << n) - 1;
  for (int mask = 1; mask <= (int)(m); ++mask) {
    f[mask] = 1;
    int submask = (mask - 1) & mask;
    while (submask) {
      long long p = 1;
      for (int i = 0; i < (int)(n); ++i) {
        if (submask >> i & 1) {
          p = p * h[i][mask ^ submask] % mod;
        }
      }
      f[mask] = (f[mask] - p * f[submask] + mod * mod) % mod;
      submask = (submask - 1) & mask;
    }
  }
  long long ans = 0;
  for (int mask = 1; mask <= (int)(m); ++mask) {
    int cnt = __builtin_popcount(mask);
    long long p = 1;
    for (int i = 0; i < (int)(n); ++i) {
      if (mask >> i & 1) {
        p = p * h[i][m ^ mask] % mod;
      }
    }
    ans = (ans + cnt * f[mask] % mod * p) % mod;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```