#include <bits/stdc++.h>
using namespace std;
const int MAXN = 14;
const int MOD = 1e9 + 7;
int N;
int a[MAXN];
int prob[MAXN][1 << MAXN];
int dp[1 << MAXN];
void load() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", a + i);
}
inline int add(int x, int y) {
  x += y;
  if (x >= MOD) return x - MOD;
  if (x < 0) return x + MOD;
  return x;
}
inline int mul(int x, int y) { return (long long)x * y % MOD; }
int pot(int x, int y) {
  int res = 1;
  for (; y; y /= 2) {
    if (y % 2) res = mul(res, x);
    x = mul(x, x);
  }
  return res;
}
int get_trans(int submask, int mask) {
  int res = dp[submask];
  for (int i = 0; i < N; i++)
    if (submask >> i & 1) res = mul(res, prob[i][mask ^ submask]);
  return res;
}
int solve() {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < 1 << N; j++) {
      prob[i][j] = 1;
      for (int k = 0; k < N; k++)
        if (j >> k & 1)
          prob[i][j] = mul(prob[i][j], mul(a[i], pot(a[i] + a[k], MOD - 2)));
    }
  for (int i = 1; i < 1 << N; i++) {
    dp[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i)
      dp[i] = add(dp[i], -get_trans(j, i));
  }
  int sol = 0;
  for (int i = 1; i < 1 << N; i++)
    sol = add(sol, mul(__builtin_popcount(i), get_trans(i, (1 << N) - 1)));
  return sol;
}
int main() {
  load();
  printf("%d\n", solve());
  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 namespace std;
const int MAXN = 14;
const int MOD = 1e9 + 7;
int N;
int a[MAXN];
int prob[MAXN][1 << MAXN];
int dp[1 << MAXN];
void load() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", a + i);
}
inline int add(int x, int y) {
  x += y;
  if (x >= MOD) return x - MOD;
  if (x < 0) return x + MOD;
  return x;
}
inline int mul(int x, int y) { return (long long)x * y % MOD; }
int pot(int x, int y) {
  int res = 1;
  for (; y; y /= 2) {
    if (y % 2) res = mul(res, x);
    x = mul(x, x);
  }
  return res;
}
int get_trans(int submask, int mask) {
  int res = dp[submask];
  for (int i = 0; i < N; i++)
    if (submask >> i & 1) res = mul(res, prob[i][mask ^ submask]);
  return res;
}
int solve() {
  for (int i = 0; i < N; i++)
    for (int j = 0; j < 1 << N; j++) {
      prob[i][j] = 1;
      for (int k = 0; k < N; k++)
        if (j >> k & 1)
          prob[i][j] = mul(prob[i][j], mul(a[i], pot(a[i] + a[k], MOD - 2)));
    }
  for (int i = 1; i < 1 << N; i++) {
    dp[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i)
      dp[i] = add(dp[i], -get_trans(j, i));
  }
  int sol = 0;
  for (int i = 1; i < 1 << N; i++)
    sol = add(sol, mul(__builtin_popcount(i), get_trans(i, (1 << N) - 1)));
  return sol;
}
int main() {
  load();
  printf("%d\n", solve());
  return 0;
}

```