#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = ' ';
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') f = -1;
  }
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
const int mod = 1e9 + 7;
int n, a[15], f[1 << 14], g[1 << 14];
int ni[15][15];
int P(int s, int t) {
  int res = 1;
  for (int i = 0; i < n; i++) {
    if (!(s & (1 << i))) continue;
    for (int j = 0; j < n; j++) {
      if (!(t & (1 << j))) continue;
      res = 1ll * res * ni[i][j] % mod;
    }
  }
  return res;
}
int fpow(int n) {
  int res = 1, m = mod - 2;
  while (m) {
    if (m & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
    m >>= 1;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) a[i] = read();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      ni[i][j] = 1ll * a[i] * fpow(a[i] + a[j]) % mod;
    }
  }
  for (int i = 0; i < (1 << n); i++) {
    long long s = (1 << n) - 1 - i;
    g[i] = P(i, s);
  }
  int ans = 0;
  for (int i = 0; i < (1 << n); i++) {
    f[i] = g[i];
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      f[i] -= 1ll * f[j] * g[i - j] % mod * fpow(P(i - j, j)) % mod;
      if (f[i] < 0) f[i] += mod;
    }
    ans += 1ll * f[i] * __builtin_popcount(i) % mod;
    if (ans > mod) ans -= mod;
  }
  cout << ans;
}

### 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;
long long read() {
  long long x = 0, f = 1;
  char ch = ' ';
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') f = -1;
  }
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
const int mod = 1e9 + 7;
int n, a[15], f[1 << 14], g[1 << 14];
int ni[15][15];
int P(int s, int t) {
  int res = 1;
  for (int i = 0; i < n; i++) {
    if (!(s & (1 << i))) continue;
    for (int j = 0; j < n; j++) {
      if (!(t & (1 << j))) continue;
      res = 1ll * res * ni[i][j] % mod;
    }
  }
  return res;
}
int fpow(int n) {
  int res = 1, m = mod - 2;
  while (m) {
    if (m & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
    m >>= 1;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) a[i] = read();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      ni[i][j] = 1ll * a[i] * fpow(a[i] + a[j]) % mod;
    }
  }
  for (int i = 0; i < (1 << n); i++) {
    long long s = (1 << n) - 1 - i;
    g[i] = P(i, s);
  }
  int ans = 0;
  for (int i = 0; i < (1 << n); i++) {
    f[i] = g[i];
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      f[i] -= 1ll * f[j] * g[i - j] % mod * fpow(P(i - j, j)) % mod;
      if (f[i] < 0) f[i] += mod;
    }
    ans += 1ll * f[i] * __builtin_popcount(i) % mod;
    if (ans > mod) ans -= mod;
  }
  cout << ans;
}

```