#include <bits/stdc++.h>
#pragma GCC optimize("O2,unroll-loops")
using namespace std;
const int inf = 1000000010;
const long long INF = 1000000000000001000LL;
const int mod = 1000000007;
const int N = 14;
int n, m, k, u, v, x, y, t, a, b, ans;
long long A[N], P[N][N];
long long tof[N][1 << N];
long long tof2[1 << 7][1 << N], tof3[1 << 7][1 << N];
long long dp[1 << N];
long long powmod(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) res = res * a % mod;
  return res;
}
long long Get(int mask1, int mask2) {
  int x = (mask1 & 127), y = ((mask1 - x) >> 7);
  return tof2[x][mask2] * tof3[y][mask2] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> A[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      P[i][j] = A[i] * powmod(A[i] + A[j], mod - 2) % mod;
  for (int i = 0; i < n; i++) {
    tof[i][0] = 1;
    for (int j = 1; j < (1 << n); j++) {
      int v = __builtin_ctz(j);
      tof[i][j] = tof[i][j ^ (1 << v)] * P[i][v] % mod;
    }
  }
  for (int i = 0; i < (1 << n); i++) tof2[0][i] = tof3[0][i] = 1;
  for (int i = 1; i < (1 << 7); i++) {
    int v = __builtin_ctz(i);
    for (int mask = 0; mask < (1 << n); mask++) {
      tof2[i][mask] = tof2[i ^ (1 << v)][mask] * tof[v][mask] % mod;
      tof3[i][mask] = tof3[i ^ (1 << v)][mask] * tof[v + 7][mask] % mod;
    }
  }
  dp[0] = 1;
  for (int mask = 1; mask < (1 << n); mask++) {
    dp[mask] = 1;
    for (int sub = (mask - 1) & mask; sub; sub = (sub - 1) & mask)
      dp[mask] = (dp[mask] - dp[sub] * Get(sub, mask ^ sub)) % mod;
    long long ted = __builtin_popcount(mask);
    ans = (ans + dp[mask] * Get(mask, (1 << n) - 1 - mask) % mod * ted) % mod;
  }
  if (ans < 0) ans += mod;
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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>
#pragma GCC optimize("O2,unroll-loops")
using namespace std;
const int inf = 1000000010;
const long long INF = 1000000000000001000LL;
const int mod = 1000000007;
const int N = 14;
int n, m, k, u, v, x, y, t, a, b, ans;
long long A[N], P[N][N];
long long tof[N][1 << N];
long long tof2[1 << 7][1 << N], tof3[1 << 7][1 << N];
long long dp[1 << N];
long long powmod(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) res = res * a % mod;
  return res;
}
long long Get(int mask1, int mask2) {
  int x = (mask1 & 127), y = ((mask1 - x) >> 7);
  return tof2[x][mask2] * tof3[y][mask2] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> A[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      P[i][j] = A[i] * powmod(A[i] + A[j], mod - 2) % mod;
  for (int i = 0; i < n; i++) {
    tof[i][0] = 1;
    for (int j = 1; j < (1 << n); j++) {
      int v = __builtin_ctz(j);
      tof[i][j] = tof[i][j ^ (1 << v)] * P[i][v] % mod;
    }
  }
  for (int i = 0; i < (1 << n); i++) tof2[0][i] = tof3[0][i] = 1;
  for (int i = 1; i < (1 << 7); i++) {
    int v = __builtin_ctz(i);
    for (int mask = 0; mask < (1 << n); mask++) {
      tof2[i][mask] = tof2[i ^ (1 << v)][mask] * tof[v][mask] % mod;
      tof3[i][mask] = tof3[i ^ (1 << v)][mask] * tof[v + 7][mask] % mod;
    }
  }
  dp[0] = 1;
  for (int mask = 1; mask < (1 << n); mask++) {
    dp[mask] = 1;
    for (int sub = (mask - 1) & mask; sub; sub = (sub - 1) & mask)
      dp[mask] = (dp[mask] - dp[sub] * Get(sub, mask ^ sub)) % mod;
    long long ted = __builtin_popcount(mask);
    ans = (ans + dp[mask] * Get(mask, (1 << n) - 1 - mask) % mod * ted) % mod;
  }
  if (ans < 0) ans += mod;
  cout << ans << "\n";
  return 0;
}

```