#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long a = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    a = a * 10 + c - 48;
    c = getchar();
  }
  return a * fh;
}
long long T, n, m;
const long long N = 2e6;
long long inv[N + N + 5];
long long f[(1 << 14 | 5)], g[(1 << 14 | 5)][15], h[(1 << 14 | 5)];
long long cnt[(1 << 14 | 5)], a[15];
long long sgn(long long n) {
  if (n & 1) return (1000000007 - 1);
  return 1;
}
signed main() {
  n = read();
  inv[1] = 1;
  for (long long i = 2; i <= N + N; ++i)
    inv[i] = (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (long long i = 1; i <= n; ++i) a[i] = read();
  for (long long st = 0; st < (1 << n); ++st) {
    cnt[st] = (cnt[st >> 1] + (st & 1));
    for (long long i = 1; i <= n; ++i) {
      if ((st >> i - 1) & 1) continue;
      g[st][i] = 1;
      for (long long j = 1; j <= n; ++j)
        if ((st >> j - 1) & 1)
          g[st][i] =
              g[st][i] * inv[a[j] + a[i]] % 1000000007 * a[j] % 1000000007;
    }
  }
  for (long long st = 1; st < (1 << n); ++st) {
    h[st] = 1;
    for (long long u = (st - 1) & st; u; u = (u - 1) & st) {
      long long A = 1, B = 1;
      for (long long j = 1; j <= n; ++j) {
        if ((u >> j - 1) & 1) A = g[st - u][j] * A % 1000000007;
        if (((st - u) >> j - 1) & 1) B = g[u][j] * B % 1000000007;
      }
      A = 0;
      h[st] = (h[st] - h[u] * (A + B) % 1000000007 + 1000000007) % 1000000007;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    for (long long j = 0; j < (1 << n); ++j) f[j] = 0;
    for (long long st = (1 << i - 1); st < (1 << n); ++st) {
      if (!((st >> i - 1) & 1)) continue;
      f[st] = (f[st] + h[st]) % 1000000007;
      long long w = ((1 << n) - 1) ^ st;
      for (long long u = w; u; u = (u - 1) & w) {
        long long to = st | u, r = 1;
        for (long long j = 1; j <= n; ++j)
          if ((u >> j - 1) & 1) r = r * g[st][j] % 1000000007;
        f[to] = (f[to] + r * f[st] % 1000000007 * h[u]) % 1000000007;
        if (!u) break;
      }
    }
    ans += f[(1 << n) - 1];
  }
  printf("%lld\n", ans % 1000000007);
  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 a = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    a = a * 10 + c - 48;
    c = getchar();
  }
  return a * fh;
}
long long T, n, m;
const long long N = 2e6;
long long inv[N + N + 5];
long long f[(1 << 14 | 5)], g[(1 << 14 | 5)][15], h[(1 << 14 | 5)];
long long cnt[(1 << 14 | 5)], a[15];
long long sgn(long long n) {
  if (n & 1) return (1000000007 - 1);
  return 1;
}
signed main() {
  n = read();
  inv[1] = 1;
  for (long long i = 2; i <= N + N; ++i)
    inv[i] = (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (long long i = 1; i <= n; ++i) a[i] = read();
  for (long long st = 0; st < (1 << n); ++st) {
    cnt[st] = (cnt[st >> 1] + (st & 1));
    for (long long i = 1; i <= n; ++i) {
      if ((st >> i - 1) & 1) continue;
      g[st][i] = 1;
      for (long long j = 1; j <= n; ++j)
        if ((st >> j - 1) & 1)
          g[st][i] =
              g[st][i] * inv[a[j] + a[i]] % 1000000007 * a[j] % 1000000007;
    }
  }
  for (long long st = 1; st < (1 << n); ++st) {
    h[st] = 1;
    for (long long u = (st - 1) & st; u; u = (u - 1) & st) {
      long long A = 1, B = 1;
      for (long long j = 1; j <= n; ++j) {
        if ((u >> j - 1) & 1) A = g[st - u][j] * A % 1000000007;
        if (((st - u) >> j - 1) & 1) B = g[u][j] * B % 1000000007;
      }
      A = 0;
      h[st] = (h[st] - h[u] * (A + B) % 1000000007 + 1000000007) % 1000000007;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    for (long long j = 0; j < (1 << n); ++j) f[j] = 0;
    for (long long st = (1 << i - 1); st < (1 << n); ++st) {
      if (!((st >> i - 1) & 1)) continue;
      f[st] = (f[st] + h[st]) % 1000000007;
      long long w = ((1 << n) - 1) ^ st;
      for (long long u = w; u; u = (u - 1) & w) {
        long long to = st | u, r = 1;
        for (long long j = 1; j <= n; ++j)
          if ((u >> j - 1) & 1) r = r * g[st][j] % 1000000007;
        f[to] = (f[to] + r * f[st] % 1000000007 * h[u]) % 1000000007;
        if (!u) break;
      }
    }
    ans += f[(1 << n) - 1];
  }
  printf("%lld\n", ans % 1000000007);
  return 0;
}

```