#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int res = 0, f = 1;
  char ch;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') f = -1;
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return res * f;
}
const int N = 14, M = 4782969, P = 1e9 + 7;
inline int qpow(int a, int b) {
  int ans = 1, base = a;
  while (b) {
    if (b & 1) ans = 1ll * ans * base % P;
    base = 1ll * base * base % P;
    b >>= 1;
  }
  return ans;
}
int rev[1 << N], cnt[1 << N], n, A, a[N];
int F[1 << N], G[1 << N], v[N][1 << N];
inline int get(int S, int T) {
  if (!S) return 1;
  if (G[S]) return G[S];
  return G[S] = 1ll * get(S & (S - 1), T) * v[rev[S & -S]][T] % P;
}
int main() {
  n = read(), A = (1 << n) - 1;
  for (int i = 0; i < n; ++i) a[i] = read(), rev[1 << i] = i;
  for (int i = 0; i < n; ++i) {
    v[i][0] = 1;
    for (int S = 1; S < (1 << n); ++S)
      v[i][S] = 1ll * v[i][S & (S - 1)] * a[i] % P *
                qpow((a[rev[S & -S]] + a[i]) % P, P - 2) % P;
  }
  for (int S = 1; S < (1 << n); ++S) {
    int tmp = 0;
    cnt[S] = cnt[S >> 1] + (S & 1);
    for (int T = (S - 1) & S; T; T = (T - 1) & S)
      (tmp += 1ll * F[T] * get(S ^ T, A ^ S) % P) %= P;
    F[S] = (get(S, A ^ S) - tmp + P) % P;
    for (int T = S; T; T = (T - 1) & S) G[T] = 0;
  }
  int res = 0;
  for (int S = 1; S < (1 << n); ++S) (res += 1ll * F[S] * cnt[S] % P) %= P;
  printf("%d", res);
  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>
using namespace std;
inline int read() {
  int res = 0, f = 1;
  char ch;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') f = -1;
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return res * f;
}
const int N = 14, M = 4782969, P = 1e9 + 7;
inline int qpow(int a, int b) {
  int ans = 1, base = a;
  while (b) {
    if (b & 1) ans = 1ll * ans * base % P;
    base = 1ll * base * base % P;
    b >>= 1;
  }
  return ans;
}
int rev[1 << N], cnt[1 << N], n, A, a[N];
int F[1 << N], G[1 << N], v[N][1 << N];
inline int get(int S, int T) {
  if (!S) return 1;
  if (G[S]) return G[S];
  return G[S] = 1ll * get(S & (S - 1), T) * v[rev[S & -S]][T] % P;
}
int main() {
  n = read(), A = (1 << n) - 1;
  for (int i = 0; i < n; ++i) a[i] = read(), rev[1 << i] = i;
  for (int i = 0; i < n; ++i) {
    v[i][0] = 1;
    for (int S = 1; S < (1 << n); ++S)
      v[i][S] = 1ll * v[i][S & (S - 1)] * a[i] % P *
                qpow((a[rev[S & -S]] + a[i]) % P, P - 2) % P;
  }
  for (int S = 1; S < (1 << n); ++S) {
    int tmp = 0;
    cnt[S] = cnt[S >> 1] + (S & 1);
    for (int T = (S - 1) & S; T; T = (T - 1) & S)
      (tmp += 1ll * F[T] * get(S ^ T, A ^ S) % P) %= P;
    F[S] = (get(S, A ^ S) - tmp + P) % P;
    for (int T = S; T; T = (T - 1) & S) G[T] = 0;
  }
  int res = 0;
  for (int S = 1; S < (1 << n); ++S) (res += 1ll * F[S] * cnt[S] % P) %= P;
  printf("%d", res);
  return 0;
}

```