#include <bits/stdc++.h>
const int maxn = 15, maxk = 1 << maxn, maxt = 2000005, mod = 1000000007;
int n, ans, maxx;
int a[maxn], inv[maxt], bitcnt[maxk], fir[maxt], pfx[maxt];
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), maxx = max(maxx, a[i]);
  inv[1] = 1;
  for (int i = 2; i <= 2 * maxx; i++)
    inv[i] = mod - 1ll * inv[mod % i] * (mod / i) % mod;
  for (int i = 1; i < (1 << n); i++) bitcnt[i] = bitcnt[i >> 1] + (i & 1);
  for (int i = 1; i < (1 << n); i++) {
    pfx[i] = 1;
    for (int j = 0; j < n; j++)
      if ((i >> j) & 1)
        for (int k = 0; k < n; k++)
          if (((i >> k) & 1) == 0)
            pfx[i] = 1ll * pfx[i] * a[j] % mod * inv[a[j] + a[k]] % mod;
    fir[i] = pfx[i];
    for (int j = (i - 1) & i; j != 0; j = (j - 1) & i) {
      int p = 1;
      for (int x = 0; x < n; x++)
        if ((j >> x) & 1)
          for (int y = 0; y < n; y++)
            if (((i >> y) & 1) && ((j >> y) & 1) == 0)
              p = 1ll * p * inv[a[x]] % mod * (a[x] + a[y]) % mod;
      fir[i] = (fir[i] - 1ll * fir[j] * pfx[i ^ j] % mod * p % mod + mod) % mod;
    }
  }
  for (int i = 1; i < (1 << n); i++)
    ans = (ans + 1ll * fir[i] * bitcnt[i] % mod) % mod;
  printf("%d\n", 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>
const int maxn = 15, maxk = 1 << maxn, maxt = 2000005, mod = 1000000007;
int n, ans, maxx;
int a[maxn], inv[maxt], bitcnt[maxk], fir[maxt], pfx[maxt];
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), maxx = max(maxx, a[i]);
  inv[1] = 1;
  for (int i = 2; i <= 2 * maxx; i++)
    inv[i] = mod - 1ll * inv[mod % i] * (mod / i) % mod;
  for (int i = 1; i < (1 << n); i++) bitcnt[i] = bitcnt[i >> 1] + (i & 1);
  for (int i = 1; i < (1 << n); i++) {
    pfx[i] = 1;
    for (int j = 0; j < n; j++)
      if ((i >> j) & 1)
        for (int k = 0; k < n; k++)
          if (((i >> k) & 1) == 0)
            pfx[i] = 1ll * pfx[i] * a[j] % mod * inv[a[j] + a[k]] % mod;
    fir[i] = pfx[i];
    for (int j = (i - 1) & i; j != 0; j = (j - 1) & i) {
      int p = 1;
      for (int x = 0; x < n; x++)
        if ((j >> x) & 1)
          for (int y = 0; y < n; y++)
            if (((i >> y) & 1) && ((j >> y) & 1) == 0)
              p = 1ll * p * inv[a[x]] % mod * (a[x] + a[y]) % mod;
      fir[i] = (fir[i] - 1ll * fir[j] * pfx[i ^ j] % mod * p % mod + mod) % mod;
    }
  }
  for (int i = 1; i < (1 << n); i++)
    ans = (ans + 1ll * fir[i] * bitcnt[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

```