#include <bits/stdc++.h>
using namespace std;
constexpr int P = 1e9 + 7;
using i64 = long long;
int norm(int x) {
  if (x < 0) {
    x += P;
  }
  if (x >= P) {
    x -= P;
  }
  return x;
}
template <class T>
T power(T a, int b) {
  T res = 1;
  for (; b; b /= 2, a *= a) {
    if (b % 2) {
      res *= a;
    }
  }
  return res;
}
struct Z {
  int x;
  Z(int x = 0) : x(norm(x)) {}
  int val() const { return x; }
  Z operator-() const { return Z(norm(P - x)); }
  Z inv() const {
    assert(x != 0);
    return power(*this, P - 2);
  }
  Z &operator*=(const Z &rhs) {
    x = i64(x) * rhs.x % P;
    return *this;
  }
  Z &operator+=(const Z &rhs) {
    x = norm(x + rhs.x);
    return *this;
  }
  Z &operator-=(const Z &rhs) {
    x = norm(x - rhs.x);
    return *this;
  }
  Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
  friend Z operator*(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res *= rhs;
    return res;
  }
  friend Z operator+(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res += rhs;
    return res;
  }
  friend Z operator-(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res -= rhs;
    return res;
  }
  friend Z operator/(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res /= rhs;
    return res;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  Z a[n];
  for (int i = 0; i < n; ++i) cin >> a[i].x;
  Z dp[1 << n], mul[1 << n][n], inner[1 << n][2];
  for (int i = 0; i < 1 << n; ++i) {
    mul[i][0] = mul[i][1] = 1;
    for (int j = 0; j < n; ++j) {
      if (i >> j & 1) mul[i][1] = mul[i][1] * a[j];
    }
    for (int j = 2; j < n; ++j) {
      mul[i][j] = mul[i][j - 1] * mul[i][1];
    }
    if (i > 0) {
      int b = 31 - __builtin_clz(i);
      inner[i][0] = inner[i ^ (1 << b)][0];
      for (int j = 0; j < b; ++j) {
        if (i >> j & 1) inner[i][0] = inner[i][0] * (a[b] + a[j]);
      }
    } else {
      inner[i][0] = 1;
    }
    inner[i][1] = 1 / inner[i][0];
  }
  for (int i = 1; i < 1 << n; ++i) {
    dp[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      dp[i] -= dp[j] * mul[j][__builtin_popcount(i ^ j)] * inner[j][0] *
               inner[i ^ j][0] * inner[i][1];
    }
  }
  Z ans = 0;
  for (int i = 1; i < 1 << n; ++i) {
    int p = (1 << n) - 1;
    ans += mul[i][__builtin_popcount(i ^ p)] * inner[i][0] * inner[i ^ p][0] *
           inner[p][1] * dp[i] * __builtin_popcount(i);
  }
  cout << ans.x << '\n';
}

### Prompt

Please formulate 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>
using namespace std;
constexpr int P = 1e9 + 7;
using i64 = long long;
int norm(int x) {
  if (x < 0) {
    x += P;
  }
  if (x >= P) {
    x -= P;
  }
  return x;
}
template <class T>
T power(T a, int b) {
  T res = 1;
  for (; b; b /= 2, a *= a) {
    if (b % 2) {
      res *= a;
    }
  }
  return res;
}
struct Z {
  int x;
  Z(int x = 0) : x(norm(x)) {}
  int val() const { return x; }
  Z operator-() const { return Z(norm(P - x)); }
  Z inv() const {
    assert(x != 0);
    return power(*this, P - 2);
  }
  Z &operator*=(const Z &rhs) {
    x = i64(x) * rhs.x % P;
    return *this;
  }
  Z &operator+=(const Z &rhs) {
    x = norm(x + rhs.x);
    return *this;
  }
  Z &operator-=(const Z &rhs) {
    x = norm(x - rhs.x);
    return *this;
  }
  Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
  friend Z operator*(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res *= rhs;
    return res;
  }
  friend Z operator+(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res += rhs;
    return res;
  }
  friend Z operator-(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res -= rhs;
    return res;
  }
  friend Z operator/(const Z &lhs, const Z &rhs) {
    Z res = lhs;
    res /= rhs;
    return res;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  Z a[n];
  for (int i = 0; i < n; ++i) cin >> a[i].x;
  Z dp[1 << n], mul[1 << n][n], inner[1 << n][2];
  for (int i = 0; i < 1 << n; ++i) {
    mul[i][0] = mul[i][1] = 1;
    for (int j = 0; j < n; ++j) {
      if (i >> j & 1) mul[i][1] = mul[i][1] * a[j];
    }
    for (int j = 2; j < n; ++j) {
      mul[i][j] = mul[i][j - 1] * mul[i][1];
    }
    if (i > 0) {
      int b = 31 - __builtin_clz(i);
      inner[i][0] = inner[i ^ (1 << b)][0];
      for (int j = 0; j < b; ++j) {
        if (i >> j & 1) inner[i][0] = inner[i][0] * (a[b] + a[j]);
      }
    } else {
      inner[i][0] = 1;
    }
    inner[i][1] = 1 / inner[i][0];
  }
  for (int i = 1; i < 1 << n; ++i) {
    dp[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      dp[i] -= dp[j] * mul[j][__builtin_popcount(i ^ j)] * inner[j][0] *
               inner[i ^ j][0] * inner[i][1];
    }
  }
  Z ans = 0;
  for (int i = 1; i < 1 << n; ++i) {
    int p = (1 << n) - 1;
    ans += mul[i][__builtin_popcount(i ^ p)] * inner[i][0] * inner[i ^ p][0] *
           inner[p][1] * dp[i] * __builtin_popcount(i);
  }
  cout << ans.x << '\n';
}

```