#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int powr(int a, long long b) {
  int x = 1 % mod;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, mod - 2); }
int p[20][20];
int a[20];
int probSCC[1 << 14];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < i; j++) {
      p[i][j] = mul(a[i], inv(a[i] + a[j]));
      p[j][i] = mul(a[j], inv(a[i] + a[j]));
    }
  }
  for (int i = 1; i < (1 << n); i++) {
    probSCC[i] = 1;
    for (int j = i; j; j = (j - 1) & i) {
      if (j == i) continue;
      int prob = 1;
      int _j = i ^ j;
      for (int u = 0; u < n; u++)
        if (j >> u & 1) {
          for (int v = 0; v < n; v++)
            if (_j >> v & 1) {
              prob = mul(prob, p[u][v]);
            }
        }
      probSCC[i] = sub(probSCC[i], mul(probSCC[j], prob));
    }
  }
  int ans = 0;
  int i = (1 << n) - 1;
  for (int j = i; j; j = (j - 1) & i) {
    int prob = probSCC[j];
    int _j = i ^ j;
    for (int u = 0; u < n; u++)
      if (j >> u & 1) {
        for (int v = 0; v < n; v++)
          if (_j >> v & 1) {
            prob = mul(prob, p[u][v]);
          }
      }
    ans = add(ans, mul(__builtin_popcount(j), prob));
  }
  cout << ans << '\n';
}

### Prompt

Your task is to create 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;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int powr(int a, long long b) {
  int x = 1 % mod;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, mod - 2); }
int p[20][20];
int a[20];
int probSCC[1 << 14];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < i; j++) {
      p[i][j] = mul(a[i], inv(a[i] + a[j]));
      p[j][i] = mul(a[j], inv(a[i] + a[j]));
    }
  }
  for (int i = 1; i < (1 << n); i++) {
    probSCC[i] = 1;
    for (int j = i; j; j = (j - 1) & i) {
      if (j == i) continue;
      int prob = 1;
      int _j = i ^ j;
      for (int u = 0; u < n; u++)
        if (j >> u & 1) {
          for (int v = 0; v < n; v++)
            if (_j >> v & 1) {
              prob = mul(prob, p[u][v]);
            }
        }
      probSCC[i] = sub(probSCC[i], mul(probSCC[j], prob));
    }
  }
  int ans = 0;
  int i = (1 << n) - 1;
  for (int j = i; j; j = (j - 1) & i) {
    int prob = probSCC[j];
    int _j = i ^ j;
    for (int u = 0; u < n; u++)
      if (j >> u & 1) {
        for (int v = 0; v < n; v++)
          if (_j >> v & 1) {
            prob = mul(prob, p[u][v]);
          }
      }
    ans = add(ans, mul(__builtin_popcount(j), prob));
  }
  cout << ans << '\n';
}

```