#include <bits/stdc++.h>
using namespace std;
const long long INFLL = 1e18;
const int INF = 1e9 + 1;
const int mod = 1e9 + 7;
long long mul(long long a, long long b) { return (a * b) % mod; }
long long add(long long a, long long b) { return ((a + b) % mod + mod) % mod; }
long long my_pow(long long n, long long m) {
  if (m == 0) return 1;
  long long now = my_pow(n, m / 2);
  if (m % 2 == 0)
    return (now * now) % mod;
  else
    return (((now * now) % mod) * n) % mod;
}
long long rev(long long a) { return my_pow(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (auto &c : a) cin >> c;
  vector<vector<long long>> h(n, vector<long long>(1 << n));
  for (int i = 0; i < n; i++) {
    for (int mask = 0; mask < (1 << n); mask++) {
      if (mask & (1 << i)) continue;
      h[i][mask] = 1;
      for (int j = 0; j < n; j++) {
        if (mask & (1 << j))
          h[i][mask] = mul(h[i][mask], mul(a[i], rev(add(a[i], a[j]))));
      }
    }
  }
  vector<long long> f(1 << n);
  long long ans = 0;
  for (int mask = 1; mask < (1 << n); mask++) {
    long long x = 1;
    for (int j = 0; j < n; j++) {
      if (!(mask & (1 << j))) continue;
      x = mul(x, h[j][((1 << n) - 1) ^ mask]);
    }
    long long y = 1;
    for (int mask1 = mask; mask1 > 0; mask1 = (mask1 - 1) & mask) {
      if (mask1 == mask) continue;
      long long z = 1;
      for (int j = 0; j < n; j++) {
        if (!(mask1 & (1 << j))) continue;
        z = mul(z, h[j][((1 << n) - 1) ^ mask]);
      }
      y = add(y, -mul(f[mask1], rev(z)));
    }
    f[mask] = mul(x, y);
    ans = add(ans, mul(f[mask], __builtin_popcount(mask)));
  }
  cout << ans;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 INFLL = 1e18;
const int INF = 1e9 + 1;
const int mod = 1e9 + 7;
long long mul(long long a, long long b) { return (a * b) % mod; }
long long add(long long a, long long b) { return ((a + b) % mod + mod) % mod; }
long long my_pow(long long n, long long m) {
  if (m == 0) return 1;
  long long now = my_pow(n, m / 2);
  if (m % 2 == 0)
    return (now * now) % mod;
  else
    return (((now * now) % mod) * n) % mod;
}
long long rev(long long a) { return my_pow(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  for (auto &c : a) cin >> c;
  vector<vector<long long>> h(n, vector<long long>(1 << n));
  for (int i = 0; i < n; i++) {
    for (int mask = 0; mask < (1 << n); mask++) {
      if (mask & (1 << i)) continue;
      h[i][mask] = 1;
      for (int j = 0; j < n; j++) {
        if (mask & (1 << j))
          h[i][mask] = mul(h[i][mask], mul(a[i], rev(add(a[i], a[j]))));
      }
    }
  }
  vector<long long> f(1 << n);
  long long ans = 0;
  for (int mask = 1; mask < (1 << n); mask++) {
    long long x = 1;
    for (int j = 0; j < n; j++) {
      if (!(mask & (1 << j))) continue;
      x = mul(x, h[j][((1 << n) - 1) ^ mask]);
    }
    long long y = 1;
    for (int mask1 = mask; mask1 > 0; mask1 = (mask1 - 1) & mask) {
      if (mask1 == mask) continue;
      long long z = 1;
      for (int j = 0; j < n; j++) {
        if (!(mask1 & (1 << j))) continue;
        z = mul(z, h[j][((1 << n) - 1) ^ mask]);
      }
      y = add(y, -mul(f[mask1], rev(z)));
    }
    f[mask] = mul(x, y);
    ans = add(ans, mul(f[mask], __builtin_popcount(mask)));
  }
  cout << ans;
  return 0;
}

```