#include <bits/stdc++.h>
using namespace std;
template <typename P, typename Q>
P CeilDiv(P _dividend, Q _divider) {
  return static_cast<P>((_dividend - 1LL + _divider) / _divider);
}
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<int>(1e9 + 7);
int Add(int x, int y) {
  x += y;
  return x < MOD ? x : x - MOD;
}
int Sub(int x, int y) {
  x -= y;
  return x >= 0 ? x : x + MOD;
}
int Mul(int x, int y) { return int(1LL * x * y % MOD); }
int FastPow(int x, int p) {
  int res = 1;
  while (p > 0) {
    if (p & 1) {
      res = Mul(res, x);
    }
    p >>= 1;
    x = Mul(x, x);
  }
  return res;
}
int Inv(int x) { return FastPow(x, MOD - 2); }
bool HasBit(int x, int b) { return (x >> b) & 1; }
void SolveTask() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  vector<vector<int>> prob(n, vector<int>(n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      prob[i][j] = Mul(arr[i], Inv(arr[i] + arr[j]));
    }
  }
  const int full_mask = 1 << n;
  vector<int> dp(full_mask, 1);
  for (int mask = 1; mask < full_mask; ++mask) {
    for (int sub = mask - 1; (sub & mask) > 0; --sub) {
      sub &= mask;
      int p = 1;
      for (int a = 0; a < n; ++a) {
        if (HasBit(sub, a)) {
          for (int b = 0; b < n; ++b) {
            if (HasBit(mask ^ sub, b)) {
              p = Mul(p, prob[a][b]);
            }
          }
        }
      }
      dp[mask] = Sub(dp[mask], Mul(dp[sub], p));
    }
  }
  vector<int> kbit(full_mask, 0);
  int ans = 0;
  for (int mask = 1; mask < full_mask; ++mask) {
    kbit[mask] = kbit[mask >> 1] + (mask & 1);
    for (int a = 0; a < n; ++a) {
      if (HasBit(mask, a)) {
        for (int b = 0; b < n; ++b) {
          if (!HasBit(mask, b)) {
            dp[mask] = Mul(dp[mask], prob[a][b]);
          }
        }
      }
    }
    ans = Add(ans, Mul(dp[mask], kbit[mask]));
  }
  cout << ans << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int test_cases = 1;
  for (int yt = 1; yt <= test_cases; ++yt) {
    SolveTask();
  }
  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;
template <typename P, typename Q>
P CeilDiv(P _dividend, Q _divider) {
  return static_cast<P>((_dividend - 1LL + _divider) / _divider);
}
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<int>(1e9 + 7);
int Add(int x, int y) {
  x += y;
  return x < MOD ? x : x - MOD;
}
int Sub(int x, int y) {
  x -= y;
  return x >= 0 ? x : x + MOD;
}
int Mul(int x, int y) { return int(1LL * x * y % MOD); }
int FastPow(int x, int p) {
  int res = 1;
  while (p > 0) {
    if (p & 1) {
      res = Mul(res, x);
    }
    p >>= 1;
    x = Mul(x, x);
  }
  return res;
}
int Inv(int x) { return FastPow(x, MOD - 2); }
bool HasBit(int x, int b) { return (x >> b) & 1; }
void SolveTask() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  vector<vector<int>> prob(n, vector<int>(n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      prob[i][j] = Mul(arr[i], Inv(arr[i] + arr[j]));
    }
  }
  const int full_mask = 1 << n;
  vector<int> dp(full_mask, 1);
  for (int mask = 1; mask < full_mask; ++mask) {
    for (int sub = mask - 1; (sub & mask) > 0; --sub) {
      sub &= mask;
      int p = 1;
      for (int a = 0; a < n; ++a) {
        if (HasBit(sub, a)) {
          for (int b = 0; b < n; ++b) {
            if (HasBit(mask ^ sub, b)) {
              p = Mul(p, prob[a][b]);
            }
          }
        }
      }
      dp[mask] = Sub(dp[mask], Mul(dp[sub], p));
    }
  }
  vector<int> kbit(full_mask, 0);
  int ans = 0;
  for (int mask = 1; mask < full_mask; ++mask) {
    kbit[mask] = kbit[mask >> 1] + (mask & 1);
    for (int a = 0; a < n; ++a) {
      if (HasBit(mask, a)) {
        for (int b = 0; b < n; ++b) {
          if (!HasBit(mask, b)) {
            dp[mask] = Mul(dp[mask], prob[a][b]);
          }
        }
      }
    }
    ans = Add(ans, Mul(dp[mask], kbit[mask]));
  }
  cout << ans << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int test_cases = 1;
  for (int yt = 1; yt <= test_cases; ++yt) {
    SolveTask();
  }
  return 0;
}

```