#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 14, mod = 1e9 + 7;
int n, n1, n2, l1, l2, lim, a[14], e[14][14], f[4][128][128], dp[N], ans;
void qmo(int &x) { x += x >> 31 & mod; }
int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) res = (long long)res * a % mod;
  return res;
}
int calc(int S, int T) {
  return (long long)f[0][S & l1 - 1][T & l1 - 1] * f[1][S & l1 - 1][T >> n1] %
         mod * f[2][S >> n1][T & l1 - 1] % mod * f[3][S >> n1][T >> n1] % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  n1 = n >> 1;
  n2 = n + 1 >> 1;
  l1 = 1 << n1;
  l2 = 1 << n2;
  lim = 1 << n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    for (int j = 0; j < i; ++j) {
      e[i][j] = (long long)a[i] * ksm(a[i] + a[j], mod - 2) % mod;
      qmo(e[j][i] = mod + 1 - e[i][j]);
    }
  }
  for (int S = 0; S < l1; ++S)
    for (int T = 0; T < l1; ++T)
      if (!(S & T)) {
        f[0][S][T] = 1;
        for (int i = 0; i < n1; ++i)
          if (S >> i & 1)
            for (int j = 0; j < n1; ++j)
              if (T >> j & 1)
                f[0][S][T] = (long long)f[0][S][T] * e[i][j] % mod;
      }
  for (int S = 0; S < l2; ++S)
    for (int T = 0; T < l2; ++T)
      if (!(S & T)) {
        f[3][S][T] = 1;
        for (int i = 0; i < n2; ++i)
          if (S >> i & 1)
            for (int j = 0; j < n2; ++j)
              if (T >> j & 1)
                f[3][S][T] = (long long)f[3][S][T] * e[i + n1][j + n1] % mod;
      }
  for (int S = 0; S < l1; ++S)
    for (int T = 0; T < l2; ++T) {
      f[1][S][T] = 1;
      for (int i = 0; i < n1; ++i)
        if (S >> i & 1)
          for (int j = 0; j < n2; ++j)
            if (T >> j & 1)
              f[1][S][T] = (long long)f[1][S][T] * e[i][j + n1] % mod;
    }
  for (int S = 0; S < l2; ++S)
    for (int T = 0; T < l1; ++T) {
      f[2][S][T] = 1;
      for (int i = 0; i < n2; ++i)
        if (S >> i & 1)
          for (int j = 0; j < n1; ++j)
            if (T >> j & 1)
              f[2][S][T] = (long long)f[2][S][T] * e[i + n1][j] % mod;
    }
  for (int i = 1; i <= n; ++i) {
    for (int S = 1; S < lim; ++S)
      if (__builtin_popcount(S) == i) {
        dp[S] = 1;
        for (int T = S & S - 1; T; T = T - 1 & S)
          qmo(dp[S] -= (long long)dp[T] * calc(T, S - T) % mod);
        ans = (ans + (long long)dp[S] * i % mod * calc(S, lim - 1 - S)) % mod;
      }
  }
  printf("%d\n", ans);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

<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 int N = 1 << 14, mod = 1e9 + 7;
int n, n1, n2, l1, l2, lim, a[14], e[14][14], f[4][128][128], dp[N], ans;
void qmo(int &x) { x += x >> 31 & mod; }
int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) res = (long long)res * a % mod;
  return res;
}
int calc(int S, int T) {
  return (long long)f[0][S & l1 - 1][T & l1 - 1] * f[1][S & l1 - 1][T >> n1] %
         mod * f[2][S >> n1][T & l1 - 1] % mod * f[3][S >> n1][T >> n1] % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  n1 = n >> 1;
  n2 = n + 1 >> 1;
  l1 = 1 << n1;
  l2 = 1 << n2;
  lim = 1 << n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    for (int j = 0; j < i; ++j) {
      e[i][j] = (long long)a[i] * ksm(a[i] + a[j], mod - 2) % mod;
      qmo(e[j][i] = mod + 1 - e[i][j]);
    }
  }
  for (int S = 0; S < l1; ++S)
    for (int T = 0; T < l1; ++T)
      if (!(S & T)) {
        f[0][S][T] = 1;
        for (int i = 0; i < n1; ++i)
          if (S >> i & 1)
            for (int j = 0; j < n1; ++j)
              if (T >> j & 1)
                f[0][S][T] = (long long)f[0][S][T] * e[i][j] % mod;
      }
  for (int S = 0; S < l2; ++S)
    for (int T = 0; T < l2; ++T)
      if (!(S & T)) {
        f[3][S][T] = 1;
        for (int i = 0; i < n2; ++i)
          if (S >> i & 1)
            for (int j = 0; j < n2; ++j)
              if (T >> j & 1)
                f[3][S][T] = (long long)f[3][S][T] * e[i + n1][j + n1] % mod;
      }
  for (int S = 0; S < l1; ++S)
    for (int T = 0; T < l2; ++T) {
      f[1][S][T] = 1;
      for (int i = 0; i < n1; ++i)
        if (S >> i & 1)
          for (int j = 0; j < n2; ++j)
            if (T >> j & 1)
              f[1][S][T] = (long long)f[1][S][T] * e[i][j + n1] % mod;
    }
  for (int S = 0; S < l2; ++S)
    for (int T = 0; T < l1; ++T) {
      f[2][S][T] = 1;
      for (int i = 0; i < n2; ++i)
        if (S >> i & 1)
          for (int j = 0; j < n1; ++j)
            if (T >> j & 1)
              f[2][S][T] = (long long)f[2][S][T] * e[i + n1][j] % mod;
    }
  for (int i = 1; i <= n; ++i) {
    for (int S = 1; S < lim; ++S)
      if (__builtin_popcount(S) == i) {
        dp[S] = 1;
        for (int T = S & S - 1; T; T = T - 1 & S)
          qmo(dp[S] -= (long long)dp[T] * calc(T, S - T) % mod);
        ans = (ans + (long long)dp[S] * i % mod * calc(S, lim - 1 - S)) % mod;
      }
  }
  printf("%d\n", ans);
}

```