#include <bits/stdc++.h>
using namespace std;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
long long a[15];
long long dp[(1 << 14) + 1];
long long b1[20], b2[20];
long long inv[2020020];
long long t1, t2;
const long long mod = 1e9 + 7;
signed main() {
  long long T = 1;
  inv[1] = 1;
  inv[0] = 1;
  for (long long i = 2, iend = 2000000; i <= iend; ++i)
    inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
  while (T--) {
    n = read();
    for (long long i = 0, iend = n - 1; i <= iend; ++i) a[i] = read();
    long long m = (1 << n) - 1;
    for (long long i = 1, iend = m; i <= iend; ++i) {
      dp[i] = 1;
      for (long long j = (i - 1) & i; j; j = (j - 1) & i) {
        long long k = i - j;
        t1 = t2 = 0;
        for (long long p = 0, pend = n - 1; p <= pend; ++p) {
          if (j & (1 << p)) {
            b1[++t1] = p;
          }
          if (k & (1 << p)) {
            b2[++t2] = p;
          }
        }
        long long nw = dp[j];
        for (long long f1 = 1, f1end = t1; f1 <= f1end; ++f1) {
          for (long long f2 = 1, f2end = t2; f2 <= f2end; ++f2) {
            nw *= a[b1[f1]];
            nw %= mod;
            nw *= inv[a[b1[f1]] + a[b2[f2]]];
            nw %= mod;
          }
        }
        dp[i] += mod, dp[i] -= nw, dp[i] %= mod;
      }
    }
    long long sum = 0;
    for (long long i = 1, iend = m; i <= iend; ++i) {
      t1 = t2 = 0;
      for (long long p = 0, pend = n - 1; p <= pend; ++p) {
        if (i & (1 << p)) {
          b1[++t1] = p;
        }
        if ((m ^ i) & (1 << p)) {
          b2[++t2] = p;
        }
      }
      long long nw = dp[i];
      for (long long f1 = 1, f1end = t1; f1 <= f1end; ++f1) {
        for (long long f2 = 1, f2end = t2; f2 <= f2end; ++f2) {
          nw *= a[b1[f1]];
          nw %= mod;
          nw *= inv[a[b1[f1]] + a[b2[f2]]];
          nw %= mod;
        }
      }
      nw *= t1;
      nw %= mod;
      sum += nw;
      sum %= mod;
    }
    cout << sum;
  }
}

### 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;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
long long a[15];
long long dp[(1 << 14) + 1];
long long b1[20], b2[20];
long long inv[2020020];
long long t1, t2;
const long long mod = 1e9 + 7;
signed main() {
  long long T = 1;
  inv[1] = 1;
  inv[0] = 1;
  for (long long i = 2, iend = 2000000; i <= iend; ++i)
    inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
  while (T--) {
    n = read();
    for (long long i = 0, iend = n - 1; i <= iend; ++i) a[i] = read();
    long long m = (1 << n) - 1;
    for (long long i = 1, iend = m; i <= iend; ++i) {
      dp[i] = 1;
      for (long long j = (i - 1) & i; j; j = (j - 1) & i) {
        long long k = i - j;
        t1 = t2 = 0;
        for (long long p = 0, pend = n - 1; p <= pend; ++p) {
          if (j & (1 << p)) {
            b1[++t1] = p;
          }
          if (k & (1 << p)) {
            b2[++t2] = p;
          }
        }
        long long nw = dp[j];
        for (long long f1 = 1, f1end = t1; f1 <= f1end; ++f1) {
          for (long long f2 = 1, f2end = t2; f2 <= f2end; ++f2) {
            nw *= a[b1[f1]];
            nw %= mod;
            nw *= inv[a[b1[f1]] + a[b2[f2]]];
            nw %= mod;
          }
        }
        dp[i] += mod, dp[i] -= nw, dp[i] %= mod;
      }
    }
    long long sum = 0;
    for (long long i = 1, iend = m; i <= iend; ++i) {
      t1 = t2 = 0;
      for (long long p = 0, pend = n - 1; p <= pend; ++p) {
        if (i & (1 << p)) {
          b1[++t1] = p;
        }
        if ((m ^ i) & (1 << p)) {
          b2[++t2] = p;
        }
      }
      long long nw = dp[i];
      for (long long f1 = 1, f1end = t1; f1 <= f1end; ++f1) {
        for (long long f2 = 1, f2end = t2; f2 <= f2end; ++f2) {
          nw *= a[b1[f1]];
          nw %= mod;
          nw *= inv[a[b1[f1]] + a[b2[f2]]];
          nw %= mod;
        }
      }
      nw *= t1;
      nw %= mod;
      sum += nw;
      sum %= mod;
    }
    cout << sum;
  }
}

```