#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 17;
const ll mod = 1e9 + 7;
ll ksm(ll a, ll b) {
  ll s = 1;
  while (b) {
    if (b & 1) s = s * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return s;
}
ll n, mx, a[N], p[N][1 << N], mp[N][N], dp[1 << N];
inline ll calc(ll S, ll T) {
  ll ret = 1;
  for (int i = 0; i < n; i++) {
    if (S & 1 << i) ret = ret * p[i][T] % mod;
  }
  return ret;
}
int main() {
  cin >> n;
  mx = (1 << n) - 1;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      mp[i][j] = a[i] * ksm(a[i] + a[j], mod - 2) % mod;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int S = 0; S <= mx; S++) {
      p[i][S] = 1;
      if (S & 1 << i) continue;
      for (int j = 0; j < n; j++) {
        if (S & 1 << j) p[i][S] = p[i][S] * mp[i][j] % mod;
      }
    }
  }
  ll ans = 0;
  for (int S = 1; S <= mx; S++) {
    int cnt = 0, x = S;
    while (x) {
      cnt += x & 1;
      x >>= 1;
    }
    dp[S] = 1;
    for (int T = S & S - 1; T; T = S & T - 1) {
      dp[S] = (dp[S] - dp[T] * calc(T, S ^ T) % mod + mod) % mod;
    }
    ans = (ans + cnt * dp[S] % mod * calc(S, mx ^ S)) % mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
using ll = long long;
const ll N = 17;
const ll mod = 1e9 + 7;
ll ksm(ll a, ll b) {
  ll s = 1;
  while (b) {
    if (b & 1) s = s * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return s;
}
ll n, mx, a[N], p[N][1 << N], mp[N][N], dp[1 << N];
inline ll calc(ll S, ll T) {
  ll ret = 1;
  for (int i = 0; i < n; i++) {
    if (S & 1 << i) ret = ret * p[i][T] % mod;
  }
  return ret;
}
int main() {
  cin >> n;
  mx = (1 << n) - 1;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      mp[i][j] = a[i] * ksm(a[i] + a[j], mod - 2) % mod;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int S = 0; S <= mx; S++) {
      p[i][S] = 1;
      if (S & 1 << i) continue;
      for (int j = 0; j < n; j++) {
        if (S & 1 << j) p[i][S] = p[i][S] * mp[i][j] % mod;
      }
    }
  }
  ll ans = 0;
  for (int S = 1; S <= mx; S++) {
    int cnt = 0, x = S;
    while (x) {
      cnt += x & 1;
      x >>= 1;
    }
    dp[S] = 1;
    for (int T = S & S - 1; T; T = S & T - 1) {
      dp[S] = (dp[S] - dp[T] * calc(T, S ^ T) % mod + mod) % mod;
    }
    ans = (ans + cnt * dp[S] % mod * calc(S, mx ^ S)) % mod;
  }
  cout << ans << endl;
  return 0;
}

```