#include <bits/stdc++.h>
const int P = 1000000007;
using int64 = long long;
int norm(int x) {
  if (x < 0) x += P;
  if (x >= P) x -= P;
  return x;
}
template <class _Tp>
_Tp power(_Tp a, int b) {
  _Tp res = 1;
  for (; b; b >>= 1, a *= a) {
    if (b & 1) res *= a;
  }
  return res;
}
struct modint {
  int x;
  modint(int x = 0) : x(norm(x)) {}
  int val() const { return x; }
  modint operator-() const { return modint(norm(P - x)); }
  modint inv() const {
    assert(x != 0);
    return power(*this, P - 2);
  }
  modint &operator*=(const modint &rhs) {
    x = int64(x) * rhs.x % P;
    return *this;
  }
  modint &operator+=(const modint &rhs) {
    x = norm(x + rhs.x);
    return *this;
  }
  modint &operator-=(const modint &rhs) {
    x = norm(x - rhs.x);
    return *this;
  }
  modint &operator/=(const modint &rhs) { return *this *= rhs.inv(); }
  friend modint operator*(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res *= rhs;
    return res;
  }
  friend modint operator+(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res += rhs;
    return res;
  }
  friend modint operator-(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res -= rhs;
    return res;
  }
  friend modint operator/(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res /= rhs;
    return res;
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  int a[n];
  for (int i = 0; i < n; ++i) std::cin >> a[i];
  modint f[1 << n][n];
  modint g[1 << n];
  for (int s = 0; s < (1 << n); ++s) {
    modint v = 1;
    for (int i = 0; i < n; ++i) {
      if (s >> i & 1) v *= a[i];
    }
    f[s][0] = 1;
    for (int i = 1; i < n; ++i) f[s][i] = f[s][i - 1] * v;
    g[s] = 1;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        if ((s >> i & 1) && (s >> j & 1)) g[s] *= a[i] + a[j];
      }
    }
  }
  modint dp[1 << n];
  for (int s = 1; s < (1 << n); ++s) {
    dp[s] = g[s];
    for (int t = (s - 1) & s; t > 0; t = (t - 1) & s) {
      dp[s] -= dp[t] * g[t] * g[s ^ t] * f[t][__builtin_popcount(s ^ t)];
    }
    dp[s] /= g[s];
  }
  modint res = 0;
  int all = (1 << n) - 1;
  for (int s = 1; s < (1 << n); ++s) {
    res += dp[s] * __builtin_popcount(s) * g[s] * g[all ^ s] *
           f[s][__builtin_popcount(all ^ s)];
  }
  res /= g[all];
  std::cout << res.val() << "\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>
const int P = 1000000007;
using int64 = long long;
int norm(int x) {
  if (x < 0) x += P;
  if (x >= P) x -= P;
  return x;
}
template <class _Tp>
_Tp power(_Tp a, int b) {
  _Tp res = 1;
  for (; b; b >>= 1, a *= a) {
    if (b & 1) res *= a;
  }
  return res;
}
struct modint {
  int x;
  modint(int x = 0) : x(norm(x)) {}
  int val() const { return x; }
  modint operator-() const { return modint(norm(P - x)); }
  modint inv() const {
    assert(x != 0);
    return power(*this, P - 2);
  }
  modint &operator*=(const modint &rhs) {
    x = int64(x) * rhs.x % P;
    return *this;
  }
  modint &operator+=(const modint &rhs) {
    x = norm(x + rhs.x);
    return *this;
  }
  modint &operator-=(const modint &rhs) {
    x = norm(x - rhs.x);
    return *this;
  }
  modint &operator/=(const modint &rhs) { return *this *= rhs.inv(); }
  friend modint operator*(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res *= rhs;
    return res;
  }
  friend modint operator+(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res += rhs;
    return res;
  }
  friend modint operator-(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res -= rhs;
    return res;
  }
  friend modint operator/(const modint &lhs, const modint &rhs) {
    modint res = lhs;
    res /= rhs;
    return res;
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  int a[n];
  for (int i = 0; i < n; ++i) std::cin >> a[i];
  modint f[1 << n][n];
  modint g[1 << n];
  for (int s = 0; s < (1 << n); ++s) {
    modint v = 1;
    for (int i = 0; i < n; ++i) {
      if (s >> i & 1) v *= a[i];
    }
    f[s][0] = 1;
    for (int i = 1; i < n; ++i) f[s][i] = f[s][i - 1] * v;
    g[s] = 1;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        if ((s >> i & 1) && (s >> j & 1)) g[s] *= a[i] + a[j];
      }
    }
  }
  modint dp[1 << n];
  for (int s = 1; s < (1 << n); ++s) {
    dp[s] = g[s];
    for (int t = (s - 1) & s; t > 0; t = (t - 1) & s) {
      dp[s] -= dp[t] * g[t] * g[s ^ t] * f[t][__builtin_popcount(s ^ t)];
    }
    dp[s] /= g[s];
  }
  modint res = 0;
  int all = (1 << n) - 1;
  for (int s = 1; s < (1 << n); ++s) {
    res += dp[s] * __builtin_popcount(s) * g[s] * g[all ^ s] *
           f[s][__builtin_popcount(all ^ s)];
  }
  res /= g[all];
  std::cout << res.val() << "\n";
  return 0;
}

```