#include <bits/stdc++.h>
using namespace std;
const long long MOD = (1e9 + 7);
const long long MAX_TEAMS = (14);
const long long MAX_MASK = (5e6);
long long Powers[MAX_TEAMS + 1];
long long Strenght[MAX_TEAMS];
long long Proba[MAX_TEAMS][MAX_TEAMS];
long long ProbaMask[MAX_MASK];
long long Transpo[1 << MAX_TEAMS];
long long DP[1 << MAX_TEAMS];
long long nbTeams;
long long GetBit(long long mask, long long index) {
  return (mask / Powers[index]) % 3;
}
long long fast_expo(long long a, long long n) {
  long long ans = 1, cur = a;
  for (; n > 0; n >>= 1, cur = (cur * cur) % MOD) {
    if (n & 1) ans = (ans * cur) % MOD;
  }
  return ans;
}
void Solve() {
  scanf("%lld", &nbTeams);
  Powers[0] = 1;
  for (long long i = 0; i < nbTeams; i++) {
    scanf("%lld", &Strenght[i]);
    Powers[i + 1] = 3 * Powers[i];
  }
  long long nbMasks = Powers[nbTeams];
  for (long long i = 0; i < nbTeams; i++) {
    for (long long j = 0; j < nbTeams; j++) {
      Proba[i][j] = Strenght[i] * fast_expo(Strenght[i] + Strenght[j], MOD - 2);
      Proba[i][j] %= MOD;
    }
  }
  for (long long m = 0; m < (1 << nbTeams); m++) {
    for (long long j = 0; j < nbTeams; j++) {
      if ((m >> j) & 1) Transpo[m] += Powers[j];
    }
  }
  ProbaMask[nbMasks - 1] = 1;
  for (long long m = nbMasks - 2; m >= 0; m--) {
    long long in = 0;
    for (long long i = 0; i < nbTeams; i++) {
      if (GetBit(m, i) < 2) in = i;
    }
    ProbaMask[m] = ProbaMask[m + (2 - GetBit(m, in)) * Powers[in]];
    for (long long i = 0; i < nbTeams; i++) {
      if (GetBit(m, in) == 1 && GetBit(m, i) == 0) {
        ProbaMask[m] *= Proba[i][in];
        ProbaMask[m] %= MOD;
      }
      if (GetBit(m, in) == 0 && GetBit(m, i) == 1) {
        ProbaMask[m] *= Proba[in][i];
        ProbaMask[m] %= MOD;
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < nbTeams; i++) {
    fill_n(DP, 1 << nbTeams, 0);
    for (long long mask = 1 << i; mask < (1 << nbTeams);
         mask = (mask + 1) | (1 << i)) {
      long long newMask = Transpo[mask];
      DP[mask] = ProbaMask[newMask];
      for (long long m = (mask - 1) & mask; m > 0; m = (m - 1) & mask) {
        if (!((m >> i) & 1)) continue;
        DP[mask] -= DP[m] * ProbaMask[newMask + Transpo[m]];
        DP[mask] %= MOD;
      }
      if (DP[mask] < 0) DP[mask] += MOD;
    }
    ans += DP[(1 << nbTeams) - 1];
  }
  printf("%lld\n", ans % MOD);
  return;
}
signed main() {
  long long nbTests = 1;
  while (nbTests--) {
    Solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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 = (1e9 + 7);
const long long MAX_TEAMS = (14);
const long long MAX_MASK = (5e6);
long long Powers[MAX_TEAMS + 1];
long long Strenght[MAX_TEAMS];
long long Proba[MAX_TEAMS][MAX_TEAMS];
long long ProbaMask[MAX_MASK];
long long Transpo[1 << MAX_TEAMS];
long long DP[1 << MAX_TEAMS];
long long nbTeams;
long long GetBit(long long mask, long long index) {
  return (mask / Powers[index]) % 3;
}
long long fast_expo(long long a, long long n) {
  long long ans = 1, cur = a;
  for (; n > 0; n >>= 1, cur = (cur * cur) % MOD) {
    if (n & 1) ans = (ans * cur) % MOD;
  }
  return ans;
}
void Solve() {
  scanf("%lld", &nbTeams);
  Powers[0] = 1;
  for (long long i = 0; i < nbTeams; i++) {
    scanf("%lld", &Strenght[i]);
    Powers[i + 1] = 3 * Powers[i];
  }
  long long nbMasks = Powers[nbTeams];
  for (long long i = 0; i < nbTeams; i++) {
    for (long long j = 0; j < nbTeams; j++) {
      Proba[i][j] = Strenght[i] * fast_expo(Strenght[i] + Strenght[j], MOD - 2);
      Proba[i][j] %= MOD;
    }
  }
  for (long long m = 0; m < (1 << nbTeams); m++) {
    for (long long j = 0; j < nbTeams; j++) {
      if ((m >> j) & 1) Transpo[m] += Powers[j];
    }
  }
  ProbaMask[nbMasks - 1] = 1;
  for (long long m = nbMasks - 2; m >= 0; m--) {
    long long in = 0;
    for (long long i = 0; i < nbTeams; i++) {
      if (GetBit(m, i) < 2) in = i;
    }
    ProbaMask[m] = ProbaMask[m + (2 - GetBit(m, in)) * Powers[in]];
    for (long long i = 0; i < nbTeams; i++) {
      if (GetBit(m, in) == 1 && GetBit(m, i) == 0) {
        ProbaMask[m] *= Proba[i][in];
        ProbaMask[m] %= MOD;
      }
      if (GetBit(m, in) == 0 && GetBit(m, i) == 1) {
        ProbaMask[m] *= Proba[in][i];
        ProbaMask[m] %= MOD;
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < nbTeams; i++) {
    fill_n(DP, 1 << nbTeams, 0);
    for (long long mask = 1 << i; mask < (1 << nbTeams);
         mask = (mask + 1) | (1 << i)) {
      long long newMask = Transpo[mask];
      DP[mask] = ProbaMask[newMask];
      for (long long m = (mask - 1) & mask; m > 0; m = (m - 1) & mask) {
        if (!((m >> i) & 1)) continue;
        DP[mask] -= DP[m] * ProbaMask[newMask + Transpo[m]];
        DP[mask] %= MOD;
      }
      if (DP[mask] < 0) DP[mask] += MOD;
    }
    ans += DP[(1 << nbTeams) - 1];
  }
  printf("%lld\n", ans % MOD);
  return;
}
signed main() {
  long long nbTests = 1;
  while (nbTests--) {
    Solve();
  }
  return 0;
}

```