#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &FF) {
  int RR = 1;
  FF = 0;
  char CH = getchar();
  for (; !isdigit(CH); CH = getchar())
    if (CH == '-') RR = -RR;
  for (; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
  FF *= RR;
}
inline void file(string str) {
  freopen((str + ".in").c_str(), "r", stdin);
  freopen((str + ".out").c_str(), "w", stdout);
}
const long long mod = 1e9 + 7;
const int N = 15, M = 5e4 + 10;
int n, ai[N];
long long g[M][N], f[M], flag[M];
long long Qpow(long long x, long long k) {
  if (k <= 1) return k ? x : 1;
  long long h = Qpow(x, k / 2);
  h = h * h % mod;
  return (k & 1) ? h * x % mod : h;
}
long long suanf(int i) {
  long long p = 0;
  if (flag[i]) return f[i];
  flag[i] = 1;
  for (int j = (i - 1) & i; j; j = (j - 1) & i) {
    if (!j) break;
    long long c = 1;
    for (int k = 1; k <= n; k++)
      if ((j >> (k - 1)) & 1) (c *= g[i ^ j][k]) %= mod;
    p += c * suanf(j) % mod;
    p %= mod;
  }
  f[i] = (1 - p + mod) % mod;
  return f[i];
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(ai[i]);
  for (int i = 0; i <= (1 << n) - 1; i++)
    for (int j = 1; j <= n; j++) {
      g[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if ((i >> (k - 1)) & 1)
          (g[i][j] *= ai[j] * Qpow(ai[j] + ai[k], mod - 2) % mod) %= mod;
    }
  long long res = 0;
  for (int i = 1; i <= (1 << n) - 1; i++) {
    long long c = 1, t = 0;
    for (int j = 1; j <= n; j++)
      if ((i >> (j - 1)) & 1) (c *= g[((1 << n) - 1) ^ i][j]) %= mod, t++;
    res += suanf(i) * c % mod * t % mod;
    res %= mod;
  }
  cout << res << endl;
  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;
template <typename T>
inline void read(T &FF) {
  int RR = 1;
  FF = 0;
  char CH = getchar();
  for (; !isdigit(CH); CH = getchar())
    if (CH == '-') RR = -RR;
  for (; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
  FF *= RR;
}
inline void file(string str) {
  freopen((str + ".in").c_str(), "r", stdin);
  freopen((str + ".out").c_str(), "w", stdout);
}
const long long mod = 1e9 + 7;
const int N = 15, M = 5e4 + 10;
int n, ai[N];
long long g[M][N], f[M], flag[M];
long long Qpow(long long x, long long k) {
  if (k <= 1) return k ? x : 1;
  long long h = Qpow(x, k / 2);
  h = h * h % mod;
  return (k & 1) ? h * x % mod : h;
}
long long suanf(int i) {
  long long p = 0;
  if (flag[i]) return f[i];
  flag[i] = 1;
  for (int j = (i - 1) & i; j; j = (j - 1) & i) {
    if (!j) break;
    long long c = 1;
    for (int k = 1; k <= n; k++)
      if ((j >> (k - 1)) & 1) (c *= g[i ^ j][k]) %= mod;
    p += c * suanf(j) % mod;
    p %= mod;
  }
  f[i] = (1 - p + mod) % mod;
  return f[i];
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(ai[i]);
  for (int i = 0; i <= (1 << n) - 1; i++)
    for (int j = 1; j <= n; j++) {
      g[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if ((i >> (k - 1)) & 1)
          (g[i][j] *= ai[j] * Qpow(ai[j] + ai[k], mod - 2) % mod) %= mod;
    }
  long long res = 0;
  for (int i = 1; i <= (1 << n) - 1; i++) {
    long long c = 1, t = 0;
    for (int j = 1; j <= n; j++)
      if ((i >> (j - 1)) & 1) (c *= g[((1 << n) - 1) ^ i][j]) %= mod, t++;
    res += suanf(i) * c % mod * t % mod;
    res %= mod;
  }
  cout << res << endl;
  return 0;
}

```