#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = (ll)1e9 + 7;
ll mAdd(ll a, ll b) {
  if (a + b >= MOD)
    return a + b - MOD;
  else
    return a + b;
}
ll mSub(ll a, ll b) {
  if (a < b)
    return a + MOD - b;
  else
    return a - b;
}
ll modPow(ll a, ll b) {
  if (b & 1) return (a * modPow(a, b ^ 1)) % MOD;
  if (b == 0) return 1;
  return modPow(a * a % MOD, b >> 1);
}
const int N = 14;
ll winp[N][N];
ll mwinp[N][1 << N];
ll sconp[1 << N];
ll winAll(ll mask, ll sub) {
  ll res = 1;
  for (int i = 0; (1 << i) <= sub; ++i) {
    if (sub & (1 << i)) res = (res * mwinp[i][mask ^ sub]) % MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<ll> as(n);
  for (ll& a : as) cin >> a;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      winp[i][j] = (as[i] * modPow(mAdd(as[i], as[j]), MOD - 2)) % MOD;
    }
    for (int mask = 0; mask < (1 << n); ++mask) {
      ll res = 1;
      for (int j = 0; j < n; ++j) {
        if (mask & (1 << j)) res = (res * winp[i][j]) % MOD;
      }
      mwinp[i][mask] = res;
    }
  }
  ll ans = 0;
  ll full = (1 << n) - 1;
  for (int mask = 1; mask < (1 << n); ++mask) {
    sconp[mask] = 1;
    for (int sub = (mask - 1) & mask; sub; sub = (sub - 1) & mask) {
      ll prob = (sconp[sub] * winAll(mask, sub)) % MOD;
      sconp[mask] = mSub(sconp[mask], prob);
    }
    ll mult = (sconp[mask] * winAll(full, mask)) % MOD;
    ans = (ans + __builtin_popcount(mask) * mult) % MOD;
  }
  cout << ans << '\n';
}

### Prompt

Generate 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;
using ll = long long;
const ll MOD = (ll)1e9 + 7;
ll mAdd(ll a, ll b) {
  if (a + b >= MOD)
    return a + b - MOD;
  else
    return a + b;
}
ll mSub(ll a, ll b) {
  if (a < b)
    return a + MOD - b;
  else
    return a - b;
}
ll modPow(ll a, ll b) {
  if (b & 1) return (a * modPow(a, b ^ 1)) % MOD;
  if (b == 0) return 1;
  return modPow(a * a % MOD, b >> 1);
}
const int N = 14;
ll winp[N][N];
ll mwinp[N][1 << N];
ll sconp[1 << N];
ll winAll(ll mask, ll sub) {
  ll res = 1;
  for (int i = 0; (1 << i) <= sub; ++i) {
    if (sub & (1 << i)) res = (res * mwinp[i][mask ^ sub]) % MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<ll> as(n);
  for (ll& a : as) cin >> a;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      winp[i][j] = (as[i] * modPow(mAdd(as[i], as[j]), MOD - 2)) % MOD;
    }
    for (int mask = 0; mask < (1 << n); ++mask) {
      ll res = 1;
      for (int j = 0; j < n; ++j) {
        if (mask & (1 << j)) res = (res * winp[i][j]) % MOD;
      }
      mwinp[i][mask] = res;
    }
  }
  ll ans = 0;
  ll full = (1 << n) - 1;
  for (int mask = 1; mask < (1 << n); ++mask) {
    sconp[mask] = 1;
    for (int sub = (mask - 1) & mask; sub; sub = (sub - 1) & mask) {
      ll prob = (sconp[sub] * winAll(mask, sub)) % MOD;
      sconp[mask] = mSub(sconp[mask], prob);
    }
    ll mult = (sconp[mask] * winAll(full, mask)) % MOD;
    ans = (ans + __builtin_popcount(mask) * mult) % MOD;
  }
  cout << ans << '\n';
}

```