#include <bits/stdc++.h>
using std::cin;
using std::cout;
const int N = 16, N2 = 16400, mod = 1000000007;
int n, ALL;
int a[N];
int prob[N2];
int G[N][N], cr[N][N2];
long long PowerMod(long long a, int n, long long c = 1) {
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) c = c * a % mod;
  return c;
}
int cross(int A, int B) {
  int ret = 1;
  for (; A; A &= A - 1) ret = (long long)ret * cr[__builtin_ctz(A)][B] % mod;
  return ret;
}
int main() {
  int i, j, ans = 0;
  std::ios::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n, ALL = ~(-1 << n);
  for (i = 0; i < n; ++i) cin >> a[i];
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j) G[i][j] = PowerMod(a[i] + a[j], mod - 2, a[i]);
  for (i = 0; i < n; ++i) {
    *cr[i] = 1;
    for (j = 1; j <= ALL; ++j)
      cr[i][j] = (long long)cr[i][j & (j - 1)] * G[i][__builtin_ctz(j)] % mod;
  }
  for (i = 1; i <= ALL; ++i) {
    prob[i] = 1;
    for (j = i & (i - 1); j; j = (j - 1) & i)
      prob[i] = (prob[i] - (long long)prob[j] * cross(j, i - j)) % mod;
    prob[i] = (prob[i] + mod) % mod;
    ans = (ans + (long long)prob[i] * cross(i, ALL - i) % mod *
                     __builtin_popcount(i)) %
          mod;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

<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 std::cin;
using std::cout;
const int N = 16, N2 = 16400, mod = 1000000007;
int n, ALL;
int a[N];
int prob[N2];
int G[N][N], cr[N][N2];
long long PowerMod(long long a, int n, long long c = 1) {
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) c = c * a % mod;
  return c;
}
int cross(int A, int B) {
  int ret = 1;
  for (; A; A &= A - 1) ret = (long long)ret * cr[__builtin_ctz(A)][B] % mod;
  return ret;
}
int main() {
  int i, j, ans = 0;
  std::ios::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n, ALL = ~(-1 << n);
  for (i = 0; i < n; ++i) cin >> a[i];
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j) G[i][j] = PowerMod(a[i] + a[j], mod - 2, a[i]);
  for (i = 0; i < n; ++i) {
    *cr[i] = 1;
    for (j = 1; j <= ALL; ++j)
      cr[i][j] = (long long)cr[i][j & (j - 1)] * G[i][__builtin_ctz(j)] % mod;
  }
  for (i = 1; i <= ALL; ++i) {
    prob[i] = 1;
    for (j = i & (i - 1); j; j = (j - 1) & i)
      prob[i] = (prob[i] - (long long)prob[j] * cross(j, i - j)) % mod;
    prob[i] = (prob[i] + mod) % mod;
    ans = (ans + (long long)prob[i] * cross(i, ALL - i) % mod *
                     __builtin_popcount(i)) %
          mod;
  }
  cout << ans << '\n';
  return 0;
}

```