#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
const int N = 14;
const int mod = 1e9 + 7;
using namespace std;
int pw(int x, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = 1ll * res * x % mod;
    }
    n /= 2;
    x = 1ll * x * x % mod;
  }
  return res;
}
void add(int &x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int n;
int a[N];
int w[N][N];
int W[N][1 << N];
int cyc[1 << N];
unordered_map<int, int> d[1 << N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < i; j++) {
      w[i][j] = 1ll * a[i] * pw(a[i] + a[j], mod - 2) % mod;
      w[j][i] = 1ll * a[j] * pw(a[i] + a[j], mod - 2) % mod;
    }
  }
  int G = (1 << n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < G; j++) {
      W[i][j] = 1;
      for (int h = 0; h < n; h++) {
        if (j & (1 << h)) {
          W[i][j] = 1ll * W[i][j] * w[i][h] % mod;
        }
      }
    }
  }
  for (int i = 0; i < G; i++) {
    for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
      int cur = 1;
      for (int h = 0; h < n; h++) {
        if (j & (1 << h)) {
          cur = 1ll * cur * W[h][i ^ j] % mod;
        }
      }
      add(cyc[i], 1ll * cyc[j] * cur % mod);
    }
    cyc[i] = 1 - cyc[i];
    if (cyc[i] < 0) {
      cyc[i] += mod;
    }
  }
  auto f = [&](int i, int j) -> int {
    int res = cyc[i];
    for (int h = 0; h < n; h++) {
      if (i & (1 << h)) {
        res = 1ll * res * W[h][j] % mod;
      }
    }
    return res;
  };
  for (int i = 1; i < G; i++) {
    add(d[i][i], cyc[i]);
    int rem = (G - 1) ^ i, all = 0;
    for (int j = i; j > 0; j = (j - 1) & i) {
      add(all, d[i][j]);
    }
    for (int h = rem; h > 0; h = (h - 1) & rem) {
      add(d[i | h][h], 1ll * all * f(h, i) % mod);
    }
  }
  int res = 0;
  for (int i = 0; i < G; i++) {
    add(res, 1ll * __builtin_popcount(i) * d[G - 1][i] % mod);
  }
  cout << res << "\n";
}

### Prompt

Create a solution in Cpp for 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("Ofast")
#pragma GCC optimize("unroll-loops")
const int N = 14;
const int mod = 1e9 + 7;
using namespace std;
int pw(int x, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = 1ll * res * x % mod;
    }
    n /= 2;
    x = 1ll * x * x % mod;
  }
  return res;
}
void add(int &x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int n;
int a[N];
int w[N][N];
int W[N][1 << N];
int cyc[1 << N];
unordered_map<int, int> d[1 << N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < i; j++) {
      w[i][j] = 1ll * a[i] * pw(a[i] + a[j], mod - 2) % mod;
      w[j][i] = 1ll * a[j] * pw(a[i] + a[j], mod - 2) % mod;
    }
  }
  int G = (1 << n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < G; j++) {
      W[i][j] = 1;
      for (int h = 0; h < n; h++) {
        if (j & (1 << h)) {
          W[i][j] = 1ll * W[i][j] * w[i][h] % mod;
        }
      }
    }
  }
  for (int i = 0; i < G; i++) {
    for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
      int cur = 1;
      for (int h = 0; h < n; h++) {
        if (j & (1 << h)) {
          cur = 1ll * cur * W[h][i ^ j] % mod;
        }
      }
      add(cyc[i], 1ll * cyc[j] * cur % mod);
    }
    cyc[i] = 1 - cyc[i];
    if (cyc[i] < 0) {
      cyc[i] += mod;
    }
  }
  auto f = [&](int i, int j) -> int {
    int res = cyc[i];
    for (int h = 0; h < n; h++) {
      if (i & (1 << h)) {
        res = 1ll * res * W[h][j] % mod;
      }
    }
    return res;
  };
  for (int i = 1; i < G; i++) {
    add(d[i][i], cyc[i]);
    int rem = (G - 1) ^ i, all = 0;
    for (int j = i; j > 0; j = (j - 1) & i) {
      add(all, d[i][j]);
    }
    for (int h = rem; h > 0; h = (h - 1) & rem) {
      add(d[i | h][h], 1ll * all * f(h, i) % mod);
    }
  }
  int res = 0;
  for (int i = 0; i < G; i++) {
    add(res, 1ll * __builtin_popcount(i) * d[G - 1][i] % mod);
  }
  cout << res << "\n";
}

```