#include <bits/stdc++.h>
using namespace std;
const int maxS = (1 << 14), maxn = 18, maxa = 1e6 + 5;
const long long MOD = 1e9 + 7;
int n, mxS, a[maxn];
bool inS(int S, int x) { return (S >> (x - 1)) & 1; }
int chgS(int S, int x) { return S ^ (1 << (x - 1)); }
long long p[maxn][maxS], f[maxn][maxS], ans, inv[maxa << 1];
void init() {
  inv[1] = 1;
  for (int i = 2; i < (maxa << 1); i++)
    inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  for (int i = 1; i <= n; i++)
    for (int S = 1; S <= mxS; S++)
      if (!inS(S, i)) {
        p[i][S] = 1;
        for (int j = 1; j <= n; j++)
          if (inS(S, j))
            p[i][S] = p[i][S] * a[j] % MOD * inv[a[i] + a[j]] % MOD;
      }
}
long long g(int S1, int S2) {
  long long ret = 1;
  for (int i = 1; i <= n; i++)
    if (inS(S1, i)) ret = ret * p[i][S2] % MOD;
  return ret;
}
int main() {
  scanf("%d", &n), mxS = (1 << n) - 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  init();
  for (int i = 1; i <= n; i++) {
    for (int S = 1; S <= mxS; S++)
      if (inS(S, i)) {
        f[i][S] = 1;
        for (int S1 = (S - 1) & S; S1; S1 = (S1 - 1) & S)
          if (inS(S1, i))
            f[i][S] = (f[i][S] - f[i][S1] * g(S1, S ^ S1) % MOD + MOD) % MOD;
      }
    ans = (ans + f[i][mxS]) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

### 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 maxS = (1 << 14), maxn = 18, maxa = 1e6 + 5;
const long long MOD = 1e9 + 7;
int n, mxS, a[maxn];
bool inS(int S, int x) { return (S >> (x - 1)) & 1; }
int chgS(int S, int x) { return S ^ (1 << (x - 1)); }
long long p[maxn][maxS], f[maxn][maxS], ans, inv[maxa << 1];
void init() {
  inv[1] = 1;
  for (int i = 2; i < (maxa << 1); i++)
    inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  for (int i = 1; i <= n; i++)
    for (int S = 1; S <= mxS; S++)
      if (!inS(S, i)) {
        p[i][S] = 1;
        for (int j = 1; j <= n; j++)
          if (inS(S, j))
            p[i][S] = p[i][S] * a[j] % MOD * inv[a[i] + a[j]] % MOD;
      }
}
long long g(int S1, int S2) {
  long long ret = 1;
  for (int i = 1; i <= n; i++)
    if (inS(S1, i)) ret = ret * p[i][S2] % MOD;
  return ret;
}
int main() {
  scanf("%d", &n), mxS = (1 << n) - 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  init();
  for (int i = 1; i <= n; i++) {
    for (int S = 1; S <= mxS; S++)
      if (inS(S, i)) {
        f[i][S] = 1;
        for (int S1 = (S - 1) & S; S1; S1 = (S1 - 1) & S)
          if (inS(S1, i))
            f[i][S] = (f[i][S] - f[i][S1] * g(S1, S ^ S1) % MOD + MOD) % MOD;
      }
    ans = (ans + f[i][mxS]) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

```