#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
const long long mod = 1e9 + 7;
long long fac[2000010], inf[2000010], iv[2000010], a[14], f[1 << 14], F[2][15],
    ans;
long long ksm(long long x, long long y) {
  long long r = 1;
  while (y) y& 1 ? r = 1ll * r * x % mod : 0, x = 1ll * x * x % mod, y >>= 1;
  return r;
}
void ini(long long n) {
  for (long long i = fac[0] = 1; i <= n; i++)
    fac[i] = 1ll * fac[i - 1] * i % mod;
  inf[n] = ksm(fac[n], mod - 2);
  for (long long i = n - 1; ~i; i--) inf[i] = inf[i + 1] * (i + 1ll) % mod;
  for (long long i = 1; i <= n; i++) iv[i] = 1ll * inf[i] * fac[i - 1] % mod;
}
signed main() {
  ini(2e6);
  long long n = read();
  for (long long i = 0; i < n; i++) a[i] = read();
  for (long long i = 0, t = 1 << n, r; i < t; i++) {
    f[i] = 1;
    for (long long j = (i - 1) & i; j; j = (j - 1) & i) {
      F[0][0] = F[1][0] = 0, r = 1;
      for (long long k = 0; k < n; k++)
        if (i >> k & 1) F[j >> k & 1][++F[j >> k & 1][0]] = k;
      for (long long c1 = 1; c1 <= F[1][0]; c1++)
        for (long long c2 = 1; c2 <= F[0][0]; c2++)
          r = 1ll * r * a[F[1][c1]] % mod * iv[a[F[1][c1]] + a[F[0][c2]]] % mod;
      f[i] = (f[i] + mod - 1ll * f[j] * r % mod) % mod;
    }
    F[0][0] = F[1][0] = 0, r = 1;
    for (long long k = 0; k < n; k++) F[i >> k & 1][++F[i >> k & 1][0]] = k;
    for (long long c1 = 1; c1 <= F[1][0]; c1++)
      for (long long c2 = 1; c2 <= F[0][0]; c2++)
        r = 1ll * r * a[F[1][c1]] % mod * iv[a[F[1][c1]] + a[F[0][c2]]] % mod;
    ans = (ans + 1ll * r * f[i] % mod * F[1][0]) % mod;
  }
  cout << ans << '\n';
  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;
long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
const long long mod = 1e9 + 7;
long long fac[2000010], inf[2000010], iv[2000010], a[14], f[1 << 14], F[2][15],
    ans;
long long ksm(long long x, long long y) {
  long long r = 1;
  while (y) y& 1 ? r = 1ll * r * x % mod : 0, x = 1ll * x * x % mod, y >>= 1;
  return r;
}
void ini(long long n) {
  for (long long i = fac[0] = 1; i <= n; i++)
    fac[i] = 1ll * fac[i - 1] * i % mod;
  inf[n] = ksm(fac[n], mod - 2);
  for (long long i = n - 1; ~i; i--) inf[i] = inf[i + 1] * (i + 1ll) % mod;
  for (long long i = 1; i <= n; i++) iv[i] = 1ll * inf[i] * fac[i - 1] % mod;
}
signed main() {
  ini(2e6);
  long long n = read();
  for (long long i = 0; i < n; i++) a[i] = read();
  for (long long i = 0, t = 1 << n, r; i < t; i++) {
    f[i] = 1;
    for (long long j = (i - 1) & i; j; j = (j - 1) & i) {
      F[0][0] = F[1][0] = 0, r = 1;
      for (long long k = 0; k < n; k++)
        if (i >> k & 1) F[j >> k & 1][++F[j >> k & 1][0]] = k;
      for (long long c1 = 1; c1 <= F[1][0]; c1++)
        for (long long c2 = 1; c2 <= F[0][0]; c2++)
          r = 1ll * r * a[F[1][c1]] % mod * iv[a[F[1][c1]] + a[F[0][c2]]] % mod;
      f[i] = (f[i] + mod - 1ll * f[j] * r % mod) % mod;
    }
    F[0][0] = F[1][0] = 0, r = 1;
    for (long long k = 0; k < n; k++) F[i >> k & 1][++F[i >> k & 1][0]] = k;
    for (long long c1 = 1; c1 <= F[1][0]; c1++)
      for (long long c2 = 1; c2 <= F[0][0]; c2++)
        r = 1ll * r * a[F[1][c1]] % mod * iv[a[F[1][c1]] + a[F[0][c2]]] % mod;
    ans = (ans + 1ll * r * f[i] % mod * F[1][0]) % mod;
  }
  cout << ans << '\n';
  return 0;
}

```