#include <bits/stdc++.h>
using namespace std;
const long long N = 1000006, INF = 1e18, P = 1e9 + 7;
long long q = 1, n, m, x, y, z, l, r, k, a[N], b[N], dp[N], o[20][20], ans, w,
          e;
long long dvv(long long a, long long b) {
  long long p = P - 2, B = 1;
  while (p) {
    if (p % 2) {
      p--;
      (B *= b) %= P;
    } else {
      p >>= 1;
      (b *= b) %= P;
    }
  }
  return (a * B) % P;
}
long long aaa(int i, int e) {
  z = 1;
  for (int x = 0; x < n; x++)
    if (i & (1 << x))
      for (int y = 0; y < n; y++)
        if (!(i & (1 << y)) && (e & (1 << y))) (z *= o[x][y]) %= P;
  return (dp[i] * z) % P;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) o[i][j] = dvv(a[i], (a[i] + a[j]));
  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[i] - aaa(j, i) + P) % P;
    (ans += aaa(i, (1 << n) - 1) * __builtin_popcount(i)) %= P;
  }
  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;
const long long N = 1000006, INF = 1e18, P = 1e9 + 7;
long long q = 1, n, m, x, y, z, l, r, k, a[N], b[N], dp[N], o[20][20], ans, w,
          e;
long long dvv(long long a, long long b) {
  long long p = P - 2, B = 1;
  while (p) {
    if (p % 2) {
      p--;
      (B *= b) %= P;
    } else {
      p >>= 1;
      (b *= b) %= P;
    }
  }
  return (a * B) % P;
}
long long aaa(int i, int e) {
  z = 1;
  for (int x = 0; x < n; x++)
    if (i & (1 << x))
      for (int y = 0; y < n; y++)
        if (!(i & (1 << y)) && (e & (1 << y))) (z *= o[x][y]) %= P;
  return (dp[i] * z) % P;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) o[i][j] = dvv(a[i], (a[i] + a[j]));
  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[i] - aaa(j, i) + P) % P;
    (ans += aaa(i, (1 << n) - 1) * __builtin_popcount(i)) %= P;
  }
  cout << ans;
}

```