#include <bits/stdc++.h>
using namespace std;
void abc() { cout << endl; }
template <typename T, typename... U>
void abc(T a, U... b) {
  cout << a << ' ', abc(b...);
}
template <typename T>
void printv(T l, T r) {
  while (l != r) cout << *l << " \n"[++l == r];
}
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
template <typename T>
ostream& operator<<(ostream& o, vector<T> a) {
  bool is = false;
  for (T i : a) {
    o << (is ? ' ' : '{'), is = true, o << i;
  }
  return o << '}';
}
template <typename T>
struct vv : vector<vector<T>> {
  vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
  vv() {}
};
template <typename T>
struct vvv : vector<vv<T>> {
  vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
  vvv() {}
};
const int mod = 1e9 + 7, N = 200001, logN = 20, K = 80000;
long long int modpow(long long int a, int b) {
  long long int ans = 1;
  for (; b; b >>= 1, a = a * a % mod) {
    if (b & 1) ans = ans * a % mod;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vv<long long int> inv(n, n, 0);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      inv[i][j] = modpow((a[i] + a[j]) % mod, mod - 2);
    }
  vv<long long int> pre(1 << n, n, 1);
  vv<long long int> pw(n, n + 1, 1);
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= n; ++j) {
      pw[i][j] = pw[i][j - 1] * a[i] % mod;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int s = 0; s < 1 << n; ++s) {
      for (int j = 0; j < n; ++j)
        if (s >> j & 1) {
          pre[s][i] = pre[s][i] * inv[i][j] % mod;
        }
    }
  }
  vector<long long int> dp(1 << n, 1);
  for (int s = 0; s < 1 << n; ++s) {
    for (int t = s ^ (s & (-s)); t > 0; t = (t - 1) & s) {
      long long int res = dp[t];
      for (int j = 0; j < n; ++j)
        if (t >> j & 1) {
          res = res * pre[t ^ s][j] % mod * pw[j][__builtin_popcount(t ^ s)] %
                mod;
        }
      dp[s] -= res;
      if (dp[s] < 0) dp[s] += mod;
    }
  }
  long long int ans = 0;
  for (int s = 0; s < 1 << n; ++s) {
    int k = __builtin_popcount(s);
    long long int res = dp[s];
    for (int i = 0; i < n; ++i)
      if (s >> i & 1) {
        for (int j = 0; j < n; ++j)
          if (~s >> j & 1) {
            res = res * a[i] % mod * inv[i][j] % mod;
          }
      }
    ans += res * k % mod;
    if (ans >= mod) ans -= mod;
  }
  cout << ans << endl;
}

### 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;
void abc() { cout << endl; }
template <typename T, typename... U>
void abc(T a, U... b) {
  cout << a << ' ', abc(b...);
}
template <typename T>
void printv(T l, T r) {
  while (l != r) cout << *l << " \n"[++l == r];
}
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
template <typename T>
ostream& operator<<(ostream& o, vector<T> a) {
  bool is = false;
  for (T i : a) {
    o << (is ? ' ' : '{'), is = true, o << i;
  }
  return o << '}';
}
template <typename T>
struct vv : vector<vector<T>> {
  vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
  vv() {}
};
template <typename T>
struct vvv : vector<vv<T>> {
  vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
  vvv() {}
};
const int mod = 1e9 + 7, N = 200001, logN = 20, K = 80000;
long long int modpow(long long int a, int b) {
  long long int ans = 1;
  for (; b; b >>= 1, a = a * a % mod) {
    if (b & 1) ans = ans * a % mod;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vv<long long int> inv(n, n, 0);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      inv[i][j] = modpow((a[i] + a[j]) % mod, mod - 2);
    }
  vv<long long int> pre(1 << n, n, 1);
  vv<long long int> pw(n, n + 1, 1);
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= n; ++j) {
      pw[i][j] = pw[i][j - 1] * a[i] % mod;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int s = 0; s < 1 << n; ++s) {
      for (int j = 0; j < n; ++j)
        if (s >> j & 1) {
          pre[s][i] = pre[s][i] * inv[i][j] % mod;
        }
    }
  }
  vector<long long int> dp(1 << n, 1);
  for (int s = 0; s < 1 << n; ++s) {
    for (int t = s ^ (s & (-s)); t > 0; t = (t - 1) & s) {
      long long int res = dp[t];
      for (int j = 0; j < n; ++j)
        if (t >> j & 1) {
          res = res * pre[t ^ s][j] % mod * pw[j][__builtin_popcount(t ^ s)] %
                mod;
        }
      dp[s] -= res;
      if (dp[s] < 0) dp[s] += mod;
    }
  }
  long long int ans = 0;
  for (int s = 0; s < 1 << n; ++s) {
    int k = __builtin_popcount(s);
    long long int res = dp[s];
    for (int i = 0; i < n; ++i)
      if (s >> i & 1) {
        for (int j = 0; j < n; ++j)
          if (~s >> j & 1) {
            res = res * a[i] % mod * inv[i][j] % mod;
          }
      }
    ans += res * k % mod;
    if (ans >= mod) ans -= mod;
  }
  cout << ans << endl;
}

```