#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
const int N = 14;
int a[N], n;
long long f1[N][1 << N];
int pos[1 << N], all;
unordered_map<long long, long long> f;
int lowbit(int x) { return x & -x; }
long long qmi(int a, int b) {
  long long res = 1, t = a;
  while (b) {
    if (b & 1) res = res * t % mod;
    t = t * t % mod;
    b >>= 1;
  }
  return res;
}
long long get(long long a, long long b) { return (a << n) | b; }
void show(int s) {
  for (int i = n - 1; i > 0; i--)
    if ((s >> i) & 1)
      cout << 1;
    else
      cout << 0;
}
void dfs(int u, int s1, int s2) {
  if (u == n) {
    if (!f.count(get(s1 - lowbit(s1), s2)) && s1) {
      show(s1 - lowbit(s1));
      cout << endl;
    }
    if (s1 == 0) {
      f[get(s1, s2)] = 1;
      return;
    }
    long long t = f[get(s1 - lowbit(s1), s2)];
    t = t * f1[pos[lowbit(s1)]][s2] % mod;
    f[get(s1, s2)] = t;
    return;
  }
  dfs(u + 1, s1, s2);
  dfs(u + 1, s1, s2 | (1 << u));
  dfs(u + 1, s1 | (1 << u), s2);
}
long long sub[1 << N], g[1 << N];
long long cal(int s) {
  vector<int> num;
  for (int i = 0; i < n; i++)
    if ((s >> i) & 1) {
      num.push_back(i);
    }
  long long res = 0;
  for (int i = 1; i < (1ll << num.size()) - 1; i++) {
    sub[i] = sub[i - lowbit(i)] | (1 << num[pos[lowbit(i)]]);
    res += f[get(sub[i], s - sub[i])] * g[sub[i]] % mod;
    res %= mod;
  }
  return res;
}
int main() {
  cin >> n;
  all = (1 << n) - 1;
  for (int i = 0; i < n; i++) pos[1 << i] = i;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    f1[i][0] = 1;
    for (int j = 1; j < 1 << n; j++) {
      int t = lowbit(j);
      f1[i][j] =
          f1[i][j - t] * a[i] % mod * qmi(a[i] + a[pos[t]], mod - 2) % mod;
    }
  }
  dfs(0, 0, 0);
  long long res = 0;
  for (int s = 1; s < 1 << n; s++) {
    int c = __builtin_popcount(s);
    g[s] = (1 - cal(s) + mod) % mod;
    res += c * f[get(s, all - s)] % mod * g[s] % mod;
    res %= mod;
  }
  cout << res << endl;
}

### Prompt

Create a solution in cpp for 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 mod = 1e9 + 7;
const int N = 14;
int a[N], n;
long long f1[N][1 << N];
int pos[1 << N], all;
unordered_map<long long, long long> f;
int lowbit(int x) { return x & -x; }
long long qmi(int a, int b) {
  long long res = 1, t = a;
  while (b) {
    if (b & 1) res = res * t % mod;
    t = t * t % mod;
    b >>= 1;
  }
  return res;
}
long long get(long long a, long long b) { return (a << n) | b; }
void show(int s) {
  for (int i = n - 1; i > 0; i--)
    if ((s >> i) & 1)
      cout << 1;
    else
      cout << 0;
}
void dfs(int u, int s1, int s2) {
  if (u == n) {
    if (!f.count(get(s1 - lowbit(s1), s2)) && s1) {
      show(s1 - lowbit(s1));
      cout << endl;
    }
    if (s1 == 0) {
      f[get(s1, s2)] = 1;
      return;
    }
    long long t = f[get(s1 - lowbit(s1), s2)];
    t = t * f1[pos[lowbit(s1)]][s2] % mod;
    f[get(s1, s2)] = t;
    return;
  }
  dfs(u + 1, s1, s2);
  dfs(u + 1, s1, s2 | (1 << u));
  dfs(u + 1, s1 | (1 << u), s2);
}
long long sub[1 << N], g[1 << N];
long long cal(int s) {
  vector<int> num;
  for (int i = 0; i < n; i++)
    if ((s >> i) & 1) {
      num.push_back(i);
    }
  long long res = 0;
  for (int i = 1; i < (1ll << num.size()) - 1; i++) {
    sub[i] = sub[i - lowbit(i)] | (1 << num[pos[lowbit(i)]]);
    res += f[get(sub[i], s - sub[i])] * g[sub[i]] % mod;
    res %= mod;
  }
  return res;
}
int main() {
  cin >> n;
  all = (1 << n) - 1;
  for (int i = 0; i < n; i++) pos[1 << i] = i;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    f1[i][0] = 1;
    for (int j = 1; j < 1 << n; j++) {
      int t = lowbit(j);
      f1[i][j] =
          f1[i][j - t] * a[i] % mod * qmi(a[i] + a[pos[t]], mod - 2) % mod;
    }
  }
  dfs(0, 0, 0);
  long long res = 0;
  for (int s = 1; s < 1 << n; s++) {
    int c = __builtin_popcount(s);
    g[s] = (1 - cal(s) + mod) % mod;
    res += c * f[get(s, all - s)] % mod * g[s] % mod;
    res %= mod;
  }
  cout << res << endl;
}

```