#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
template <class T>
void Min(T &a, const T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, const T b) {
  if (a < b) a = b;
}
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 14;
const long long mod = 1e9 + 7;
int n;
long long p[N][N], ar[N], f[1 << N];
long long pm(long long a, long long b) {
  long long ret = 1;
  for (a %= mod; b; b >>= 1, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      p[i][j] = ar[i] * pm(ar[i] + ar[j], mod - 2) % mod;
    }
  for (int s = 1; s < 1 << n; s++) {
    f[s] = 1;
    long long sub = 0;
    for (int t = s - 1 & s; t; t = t - 1 & s) {
      long long a = f[t];
      for (int i = 0; i < n; i++)
        if (t & 1 << i) {
          for (int j = 0; j < n; j++)
            if ((t & 1 << j) == 0 && s & 1 << j) {
              a = a * p[i][j] % mod;
            }
        }
      sub = (sub + a) % mod;
    }
    f[s] = (f[s] - sub + mod) % mod;
  }
  long long ans = 0;
  for (int s = 0; s < 1 << n; s++) {
    long long cnt = 0, add = f[s];
    for (int i = 0; i < n; i++)
      if (s & 1 << i) {
        cnt++;
        for (int j = 0; j < n; j++)
          if ((s & 1 << j) == 0) {
            add = add * p[i][j] % mod;
          }
      }
    add = add * cnt % mod;
    ans = (ans + add) % mod;
  }
  cout << ans;
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
template <class T>
void Min(T &a, const T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, const T b) {
  if (a < b) a = b;
}
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 14;
const long long mod = 1e9 + 7;
int n;
long long p[N][N], ar[N], f[1 << N];
long long pm(long long a, long long b) {
  long long ret = 1;
  for (a %= mod; b; b >>= 1, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      p[i][j] = ar[i] * pm(ar[i] + ar[j], mod - 2) % mod;
    }
  for (int s = 1; s < 1 << n; s++) {
    f[s] = 1;
    long long sub = 0;
    for (int t = s - 1 & s; t; t = t - 1 & s) {
      long long a = f[t];
      for (int i = 0; i < n; i++)
        if (t & 1 << i) {
          for (int j = 0; j < n; j++)
            if ((t & 1 << j) == 0 && s & 1 << j) {
              a = a * p[i][j] % mod;
            }
        }
      sub = (sub + a) % mod;
    }
    f[s] = (f[s] - sub + mod) % mod;
  }
  long long ans = 0;
  for (int s = 0; s < 1 << n; s++) {
    long long cnt = 0, add = f[s];
    for (int i = 0; i < n; i++)
      if (s & 1 << i) {
        cnt++;
        for (int j = 0; j < n; j++)
          if ((s & 1 << j) == 0) {
            add = add * p[i][j] % mod;
          }
      }
    add = add * cnt % mod;
    ans = (ans + add) % mod;
  }
  cout << ans;
}

```