#include <bits/stdc++.h>
using namespace std;
const int N = 2e6, mod = 1e9 + 7;
int read() {
  int a = 0;
  bool opt = 0;
  char c = getchar();
  while (c < '0' || c > '9') opt |= c == '-', c = getchar();
  while (c >= '0' && c <= '9')
    a = (a << 3) + (a << 1) + (c ^ 48), c = getchar();
  return opt ? -a : a;
}
int Mod(int x) { return x < mod ? x : x - mod; }
void Add(int& x, int y) { x = Mod(x + y); }
int n, ans, s, a[20], f[1 << 20], size[1 << 20], inv[2000005];
int solve(int A, int B) {
  int ans = 1;
  for (int i = 1; i <= n; ++i) {
    if (A & (1 << i - 1)) {
      for (int j = 1; j <= n; ++j) {
        if (B & (1 << j - 1)) {
          ans = 1ll * ans * a[i] % mod * inv[a[i] + a[j]] % mod;
        }
      }
    }
  }
  return ans;
}
int main() {
  inv[1] = 1;
  for (int i = 2; i <= N; ++i)
    inv[i] = mod - 1ll * (mod / i) * inv[mod % i] % mod;
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  f[0] = 1;
  s = (1 << n) - 1;
  for (int i = 1; i < (1 << n); ++i) {
    f[i] = 1;
    size[i] = size[i >> 1] + (i & 1);
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      Add(f[i], mod - 1ll * f[j] * solve(j, i - j) % mod);
    }
    Add(ans, 1ll * f[i] * size[i] % mod * solve(i, s - i) % mod);
  }
  cout << ans << "\n";
  return 0;
}

### 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;
const int N = 2e6, mod = 1e9 + 7;
int read() {
  int a = 0;
  bool opt = 0;
  char c = getchar();
  while (c < '0' || c > '9') opt |= c == '-', c = getchar();
  while (c >= '0' && c <= '9')
    a = (a << 3) + (a << 1) + (c ^ 48), c = getchar();
  return opt ? -a : a;
}
int Mod(int x) { return x < mod ? x : x - mod; }
void Add(int& x, int y) { x = Mod(x + y); }
int n, ans, s, a[20], f[1 << 20], size[1 << 20], inv[2000005];
int solve(int A, int B) {
  int ans = 1;
  for (int i = 1; i <= n; ++i) {
    if (A & (1 << i - 1)) {
      for (int j = 1; j <= n; ++j) {
        if (B & (1 << j - 1)) {
          ans = 1ll * ans * a[i] % mod * inv[a[i] + a[j]] % mod;
        }
      }
    }
  }
  return ans;
}
int main() {
  inv[1] = 1;
  for (int i = 2; i <= N; ++i)
    inv[i] = mod - 1ll * (mod / i) * inv[mod % i] % mod;
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  f[0] = 1;
  s = (1 << n) - 1;
  for (int i = 1; i < (1 << n); ++i) {
    f[i] = 1;
    size[i] = size[i >> 1] + (i & 1);
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      Add(f[i], mod - 1ll * f[j] * solve(j, i - j) % mod);
    }
    Add(ans, 1ll * f[i] * size[i] % mod * solve(i, s - i) % mod);
  }
  cout << ans << "\n";
  return 0;
}

```