#include <bits/stdc++.h>
using namespace std;
long long mulMod(long long a, long long b, long long m = 1000000007) {
  long long x = 0, y = a % m;
  while (b > 0) {
    if (b % 2 == 1) x = (x + y) % m;
    y = (y * 2) % m;
    b /= 2;
  }
  return x % m;
}
long long expMod(long long b, long long e, long long m = 1000000007) {
  if (!e) return 1;
  long long q = expMod(b, e / 2, m);
  q = mulMod(q, q, m);
  return e % 2 ? mulMod(b, q, m) : q;
}
long long invFact[20], fact[20];
long long inverso(long long x) { return expMod(x, 1000000007 - 2); }
void pre() {
  fact[0] = 1;
  for (int i = (1); i < (20 - 2); i++) fact[i] = (fact[i - 1] * i) % 1000000007;
  invFact[20 - 3] = inverso(fact[20 - 3]);
  for (int i = 20 - 3 - 1; i >= 0; i--)
    invFact[i] = (invFact[i + 1] * (i + 1)) % 1000000007;
  return;
}
long long comb(long long nn, long long k) {
  if (k < 0 || nn < k) return 0;
  return (fact[nn] * invFact[k] % 1000000007) * invFact[nn - k] % 1000000007;
}
int n;
long long win[15][15];
long long get(int mask1, int mask2) {
  long long ret = 1;
  for (int i = (0); i < (n); i++) {
    if ((mask1 >> i) & 1) {
      for (int j = (0); j < (n); j++) {
        if ((mask2 >> j) & 1) {
          ret = ((ret * win[i][j] % 1000000007) + 1000000007) % 1000000007;
          assert(ret >= 0);
        }
      }
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<long long> a(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++)
      win[i][j] = (((a[i] * inverso(a[i] + a[j])) % 1000000007) + 1000000007) %
                  1000000007;
  vector<long long> proba(1 << n, 1);
  long long rta = 0;
  for (int mask = 1; mask < (1 << n); mask++) {
    for (int submask = mask & (mask - 1); submask;
         submask = (submask - 1) & mask) {
      proba[mask] -= proba[submask] * get(submask, mask ^ submask);
      proba[mask] = ((proba[mask] % 1000000007) + 1000000007) % 1000000007;
      assert(proba[mask] >= 0);
    }
    rta += ((__builtin_popcount(mask) * proba[mask]) % 1000000007 *
            get(mask, ((1 << n) - 1) ^ mask)) %
           1000000007;
    rta = ((rta % 1000000007) + 1000000007) % 1000000007;
  }
  cout << rta << '\n';
  return 0;
}

### Prompt

Please formulate 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;
long long mulMod(long long a, long long b, long long m = 1000000007) {
  long long x = 0, y = a % m;
  while (b > 0) {
    if (b % 2 == 1) x = (x + y) % m;
    y = (y * 2) % m;
    b /= 2;
  }
  return x % m;
}
long long expMod(long long b, long long e, long long m = 1000000007) {
  if (!e) return 1;
  long long q = expMod(b, e / 2, m);
  q = mulMod(q, q, m);
  return e % 2 ? mulMod(b, q, m) : q;
}
long long invFact[20], fact[20];
long long inverso(long long x) { return expMod(x, 1000000007 - 2); }
void pre() {
  fact[0] = 1;
  for (int i = (1); i < (20 - 2); i++) fact[i] = (fact[i - 1] * i) % 1000000007;
  invFact[20 - 3] = inverso(fact[20 - 3]);
  for (int i = 20 - 3 - 1; i >= 0; i--)
    invFact[i] = (invFact[i + 1] * (i + 1)) % 1000000007;
  return;
}
long long comb(long long nn, long long k) {
  if (k < 0 || nn < k) return 0;
  return (fact[nn] * invFact[k] % 1000000007) * invFact[nn - k] % 1000000007;
}
int n;
long long win[15][15];
long long get(int mask1, int mask2) {
  long long ret = 1;
  for (int i = (0); i < (n); i++) {
    if ((mask1 >> i) & 1) {
      for (int j = (0); j < (n); j++) {
        if ((mask2 >> j) & 1) {
          ret = ((ret * win[i][j] % 1000000007) + 1000000007) % 1000000007;
          assert(ret >= 0);
        }
      }
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<long long> a(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++)
      win[i][j] = (((a[i] * inverso(a[i] + a[j])) % 1000000007) + 1000000007) %
                  1000000007;
  vector<long long> proba(1 << n, 1);
  long long rta = 0;
  for (int mask = 1; mask < (1 << n); mask++) {
    for (int submask = mask & (mask - 1); submask;
         submask = (submask - 1) & mask) {
      proba[mask] -= proba[submask] * get(submask, mask ^ submask);
      proba[mask] = ((proba[mask] % 1000000007) + 1000000007) % 1000000007;
      assert(proba[mask] >= 0);
    }
    rta += ((__builtin_popcount(mask) * proba[mask]) % 1000000007 *
            get(mask, ((1 << n) - 1) ^ mask)) %
           1000000007;
    rta = ((rta % 1000000007) + 1000000007) % 1000000007;
  }
  cout << rta << '\n';
  return 0;
}

```