#include <bits/stdc++.h>
using namespace std;
int const N = 14;
int const mod = 1e9 + 7;
long long ext_gcd(long long a, long long b, long long &x, long long &y) {
  long long t, ret;
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  ret = ext_gcd(b, a % b, x, y);
  t = x, x = y, y = t - a / b * y;
  return ret;
}
long long inv(long long a, long long b, long long c) {
  long long x, y;
  ext_gcd(a, c, x, y);
  return (x * b % c + c) % c;
}
int n;
long long a[N];
long long ff[1 << N], ga[1 << N][N + 1], gb[1 << N];
long long g(int x, int y) {
  long long da = ga[x][__builtin_popcount(y)];
  long long db = inv(gb[x ^ y], gb[x] * gb[y] % mod, mod);
  return da * db % mod;
}
long long f(int msk, int all) {
  if (ff[msk] != -1) return ff[msk];
  long long ans = g(msk, all ^ msk);
  for (int i = (msk - 1) & msk; i > 0; i = (i - 1) & msk) {
    ans -= f(i, all) * g(msk ^ i, all ^ msk);
    ans %= mod;
  }
  return ff[msk] = ans;
}
void ginit(int n) {
  for (int i = 0; i < (1 << n); ++i) {
    ga[i][0] = gb[i] = 1;
    long long z = 1;
    for (int j = 0; j < (n); ++j)
      if (i >> j & 1) z = z * a[j] % mod;
    for (int j = 1; j <= (n); ++j) ga[i][j] = ga[i][j - 1] * z % mod;
    for (int j = 0; j < (n); ++j)
      for (int k = j + 1; k < n; ++k) {
        if ((i >> j & 1) && (i >> k & 1)) {
          gb[i] = gb[i] * (a[j] + a[k]) % mod;
        }
      }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) scanf("%lld", &a[i]);
  for (int i = 0; i < (1 << n); ++i) ff[i] = -1;
  ginit(n);
  long long ans = 0;
  int all = (1 << n) - 1;
  for (int i = 1; i <= (all); ++i) {
    ans += f(i, all) * __builtin_popcount(i);
    ans %= mod;
  }
  ans = (ans + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
int const N = 14;
int const mod = 1e9 + 7;
long long ext_gcd(long long a, long long b, long long &x, long long &y) {
  long long t, ret;
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  ret = ext_gcd(b, a % b, x, y);
  t = x, x = y, y = t - a / b * y;
  return ret;
}
long long inv(long long a, long long b, long long c) {
  long long x, y;
  ext_gcd(a, c, x, y);
  return (x * b % c + c) % c;
}
int n;
long long a[N];
long long ff[1 << N], ga[1 << N][N + 1], gb[1 << N];
long long g(int x, int y) {
  long long da = ga[x][__builtin_popcount(y)];
  long long db = inv(gb[x ^ y], gb[x] * gb[y] % mod, mod);
  return da * db % mod;
}
long long f(int msk, int all) {
  if (ff[msk] != -1) return ff[msk];
  long long ans = g(msk, all ^ msk);
  for (int i = (msk - 1) & msk; i > 0; i = (i - 1) & msk) {
    ans -= f(i, all) * g(msk ^ i, all ^ msk);
    ans %= mod;
  }
  return ff[msk] = ans;
}
void ginit(int n) {
  for (int i = 0; i < (1 << n); ++i) {
    ga[i][0] = gb[i] = 1;
    long long z = 1;
    for (int j = 0; j < (n); ++j)
      if (i >> j & 1) z = z * a[j] % mod;
    for (int j = 1; j <= (n); ++j) ga[i][j] = ga[i][j - 1] * z % mod;
    for (int j = 0; j < (n); ++j)
      for (int k = j + 1; k < n; ++k) {
        if ((i >> j & 1) && (i >> k & 1)) {
          gb[i] = gb[i] * (a[j] + a[k]) % mod;
        }
      }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) scanf("%lld", &a[i]);
  for (int i = 0; i < (1 << n); ++i) ff[i] = -1;
  ginit(n);
  long long ans = 0;
  int all = (1 << n) - 1;
  for (int i = 1; i <= (all); ++i) {
    ans += f(i, all) * __builtin_popcount(i);
    ans %= mod;
  }
  ans = (ans + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

```