#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007LL;
long long FastPow(long long k, long long p) {
  if (p == 0) return 1LL;
  long long P = FastPow(k, p / 2LL);
  P = (P * P) % MOD;
  if (p % 2LL == 1LL) {
    P = (P * k) % MOD;
  }
  return P;
}
long long Inv(long long x) { return FastPow(x, MOD - 2LL); }
int n;
long long a[21];
long long probs[21][21];
inline bool bit(int k, int b) { return (k & (1 << b)) > 0; }
string tob(int k) {
  if (k == 0) return "0";
  string s;
  while (k > 0) {
    s.push_back(k % 2 + '0');
    k /= 2;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long F[100111];
long long caseProbPrec[15][100111];
int main() {
  int i, j, in, ja;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  long long ans = 0LL;
  for (int iter = 0; iter < n; iter++) {
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        probs[i][j] = (a[i] * Inv(a[i] + a[j])) % MOD;
      }
    }
    for (i = 0; i < n - 1; i++) {
      for (int sub = 0; sub < (1 << (n - 1)); sub++) {
        caseProbPrec[i][sub] = probs[i][n - 1];
        for (j = 0; j < n - 1; j++) {
          if (bit(sub, j)) {
            caseProbPrec[i][sub] *= probs[i][j];
            caseProbPrec[i][sub] %= MOD;
          }
        }
      }
    }
    F[0] = 1LL;
    for (int mask = 1; mask < (1 << (n - 1)); mask++) {
      F[mask] = 1LL;
      for (int sub = ((mask - 1) & mask); sub >= 0; sub = (sub - 1) & mask) {
        long long caseProb = F[sub];
        for (i = 0; i < n - 1; i++) {
          if (bit(mask, i) && !bit(sub, i)) {
            caseProb *= caseProbPrec[i][sub];
            caseProb %= MOD;
          }
        }
        F[mask] -= caseProb;
        if (F[mask] < 0) F[mask] += MOD;
        if (sub == 0) break;
      }
    }
    ans += F[(1 << (n - 1)) - 1];
    if (ans >= MOD) ans -= MOD;
    swap(a[iter], a[n - 1]);
  }
  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;
const long long MOD = 1000000007LL;
long long FastPow(long long k, long long p) {
  if (p == 0) return 1LL;
  long long P = FastPow(k, p / 2LL);
  P = (P * P) % MOD;
  if (p % 2LL == 1LL) {
    P = (P * k) % MOD;
  }
  return P;
}
long long Inv(long long x) { return FastPow(x, MOD - 2LL); }
int n;
long long a[21];
long long probs[21][21];
inline bool bit(int k, int b) { return (k & (1 << b)) > 0; }
string tob(int k) {
  if (k == 0) return "0";
  string s;
  while (k > 0) {
    s.push_back(k % 2 + '0');
    k /= 2;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long F[100111];
long long caseProbPrec[15][100111];
int main() {
  int i, j, in, ja;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  long long ans = 0LL;
  for (int iter = 0; iter < n; iter++) {
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        probs[i][j] = (a[i] * Inv(a[i] + a[j])) % MOD;
      }
    }
    for (i = 0; i < n - 1; i++) {
      for (int sub = 0; sub < (1 << (n - 1)); sub++) {
        caseProbPrec[i][sub] = probs[i][n - 1];
        for (j = 0; j < n - 1; j++) {
          if (bit(sub, j)) {
            caseProbPrec[i][sub] *= probs[i][j];
            caseProbPrec[i][sub] %= MOD;
          }
        }
      }
    }
    F[0] = 1LL;
    for (int mask = 1; mask < (1 << (n - 1)); mask++) {
      F[mask] = 1LL;
      for (int sub = ((mask - 1) & mask); sub >= 0; sub = (sub - 1) & mask) {
        long long caseProb = F[sub];
        for (i = 0; i < n - 1; i++) {
          if (bit(mask, i) && !bit(sub, i)) {
            caseProb *= caseProbPrec[i][sub];
            caseProb %= MOD;
          }
        }
        F[mask] -= caseProb;
        if (F[mask] < 0) F[mask] += MOD;
        if (sub == 0) break;
      }
    }
    ans += F[(1 << (n - 1)) - 1];
    if (ans >= MOD) ans -= MOD;
    swap(a[iter], a[n - 1]);
  }
  printf("%lld\n", ans);
  return 0;
}

```