#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll pwr(ll a, ll pw) {
  ll ret = 1;
  while (pw > 0) {
    if (pw & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    pw >>= 1;
  }
  return ret;
}
ll modinv(ll a) { return pwr(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  vector<ll> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<vector<ll> > prob_win(n, vector<ll>(n));
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      prob_win[i][j] = v[i] * modinv(v[i] + v[j]) % mod;
      prob_win[j][i] = v[j] * modinv(v[i] + v[j]) % mod;
    }
  }
  vector<vector<ll> > win_msk(n, vector<ll>(1 << n, 1ll));
  for (int i = 0; i < n; i++) {
    for (int msk = 0; msk < 1 << n; msk++) {
      for (int bit = 0; bit < n; bit++) {
        if (msk >> bit & 1) {
          win_msk[i][msk] = win_msk[i][msk] * prob_win[i][bit] % mod;
        }
      }
    }
  }
  ll ans = 0;
  vector<ll> dp_prob(1 << n, 1ll);
  for (int msk = 1; msk < 1 << n; msk++) {
    ll num_winners = __builtin_popcount(msk);
    for (int bit = 0; bit < n; bit++) {
      if (msk >> bit & 1) {
        dp_prob[msk] = dp_prob[msk] * win_msk[bit][(1 << n) - 1 - msk] % mod;
      }
    }
    for (int sub_winners = (msk - 1) & msk; sub_winners > 0;
         sub_winners = (sub_winners - 1) & msk) {
      int sub_losers = msk ^ sub_winners;
      ll p = dp_prob[sub_winners];
      for (int bit = 0; bit < n; bit++) {
        if (sub_losers >> bit & 1) {
          p = p * win_msk[bit][(1 << n) - 1 - msk] % mod;
        }
      }
      dp_prob[msk] = (dp_prob[msk] - p + mod) % mod;
    }
    ans = (ans + dp_prob[msk] * num_winners % mod) % mod;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll pwr(ll a, ll pw) {
  ll ret = 1;
  while (pw > 0) {
    if (pw & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    pw >>= 1;
  }
  return ret;
}
ll modinv(ll a) { return pwr(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  vector<ll> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  vector<vector<ll> > prob_win(n, vector<ll>(n));
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      prob_win[i][j] = v[i] * modinv(v[i] + v[j]) % mod;
      prob_win[j][i] = v[j] * modinv(v[i] + v[j]) % mod;
    }
  }
  vector<vector<ll> > win_msk(n, vector<ll>(1 << n, 1ll));
  for (int i = 0; i < n; i++) {
    for (int msk = 0; msk < 1 << n; msk++) {
      for (int bit = 0; bit < n; bit++) {
        if (msk >> bit & 1) {
          win_msk[i][msk] = win_msk[i][msk] * prob_win[i][bit] % mod;
        }
      }
    }
  }
  ll ans = 0;
  vector<ll> dp_prob(1 << n, 1ll);
  for (int msk = 1; msk < 1 << n; msk++) {
    ll num_winners = __builtin_popcount(msk);
    for (int bit = 0; bit < n; bit++) {
      if (msk >> bit & 1) {
        dp_prob[msk] = dp_prob[msk] * win_msk[bit][(1 << n) - 1 - msk] % mod;
      }
    }
    for (int sub_winners = (msk - 1) & msk; sub_winners > 0;
         sub_winners = (sub_winners - 1) & msk) {
      int sub_losers = msk ^ sub_winners;
      ll p = dp_prob[sub_winners];
      for (int bit = 0; bit < n; bit++) {
        if (sub_losers >> bit & 1) {
          p = p * win_msk[bit][(1 << n) - 1 - msk] % mod;
        }
      }
      dp_prob[msk] = (dp_prob[msk] - p + mod) % mod;
    }
    ans = (ans + dp_prob[msk] * num_winners % mod) % mod;
  }
  cout << ans << "\n";
  return 0;
}

```