#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int add(int x, int y) { return x + y >= MOD ? x + y - MOD : x + y; }
int sub(int x, int y) { return x >= y ? x - y : x + MOD - y; }
long long fast_pow(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * b % MOD;
    b = b * b % MOD;
    p >>= 1;
  }
  return ans;
}
long long INV(long long x) { return fast_pow(x, MOD - 2); }
int n;
int a[20];
long long p[20][20];
long long dp[1 << 14];
long long calc(int s, int t) {
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    if (s & (1 << i)) {
      for (int j = 0; j < n; j++)
        if (t & (1 << j)) ans = ans * p[i][j] % MOD;
    }
  }
  return ans;
}
int cntbit[1 << 14];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int U = (1 << n) - 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j) p[i][j] = a[i] * INV(a[i] + a[j]) % MOD;
  long long ans = 0;
  for (int i = 1; i < (1 << n); i++) {
    long long x = calc(i, U ^ i);
    dp[i] = x;
    for (int j = i; j; j = (j - 1) & i) {
      if (j == i) continue;
      dp[i] = sub(dp[i], dp[j] * calc(i ^ j, U ^ i) % MOD);
    }
    cntbit[i] = cntbit[i >> 1] + (i & 1);
    ans = add(ans, cntbit[i] * dp[i] % MOD);
  }
  printf("%lld", 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 long long MOD = 1e9 + 7;
int add(int x, int y) { return x + y >= MOD ? x + y - MOD : x + y; }
int sub(int x, int y) { return x >= y ? x - y : x + MOD - y; }
long long fast_pow(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * b % MOD;
    b = b * b % MOD;
    p >>= 1;
  }
  return ans;
}
long long INV(long long x) { return fast_pow(x, MOD - 2); }
int n;
int a[20];
long long p[20][20];
long long dp[1 << 14];
long long calc(int s, int t) {
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    if (s & (1 << i)) {
      for (int j = 0; j < n; j++)
        if (t & (1 << j)) ans = ans * p[i][j] % MOD;
    }
  }
  return ans;
}
int cntbit[1 << 14];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int U = (1 << n) - 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j) p[i][j] = a[i] * INV(a[i] + a[j]) % MOD;
  long long ans = 0;
  for (int i = 1; i < (1 << n); i++) {
    long long x = calc(i, U ^ i);
    dp[i] = x;
    for (int j = i; j; j = (j - 1) & i) {
      if (j == i) continue;
      dp[i] = sub(dp[i], dp[j] * calc(i ^ j, U ^ i) % MOD);
    }
    cntbit[i] = cntbit[i >> 1] + (i & 1);
    ans = add(ans, cntbit[i] * dp[i] % MOD);
  }
  printf("%lld", ans);
}

```