#include <bits/stdc++.h>
using namespace std;
int inv(int v) {
  return v <= 1 ? v
                : (1000000007) - (long long)inv((1000000007) % v) *
                                     ((1000000007) / v) % (1000000007);
}
struct mint {
  int v;
  mint() = default;
  mint(long long v) : v((v % (1000000007) + (1000000007)) % (1000000007)) {}
  mint& operator+=(mint a) {
    v = (v + a.v) % (1000000007);
    return *this;
  }
  mint& operator-=(mint a) {
    v = (v - a.v + (1000000007)) % (1000000007);
    return *this;
  }
  mint& operator*=(mint a) {
    v = ((long long)v * a.v) % (1000000007);
    return *this;
  }
  mint& operator/=(mint a) {
    v = ((long long)v * inv(a.v)) % (1000000007);
    return *this;
  }
  friend mint operator+(mint a, mint b) { return mint(a.v + b.v); }
  friend mint operator-(mint a, mint b) { return mint(a.v - b.v); }
  friend mint operator*(mint a, mint b) { return mint((long long)a.v * b.v); }
  friend mint operator/(mint a, mint b) {
    return mint((long long)a.v * inv(b.v));
  }
};
mint pow(mint base, long long exponent) {
  if (exponent <= 0) {
    return 1;
  }
  mint ret = 1;
  while (exponent) {
    if (exponent & 1) {
      ret *= base;
    }
    exponent >>= 1;
    base = base * base;
  }
  return ret;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  while (cin >> n) {
    vector<int> w(n);
    for (int i = 0; i < n; ++i) {
      cin >> w[i];
    }
    vector<vector<mint>> a(n, vector<mint>(n));
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        a[i][j] = mint(w[i]) / mint(w[i] + w[j]);
      }
    }
    vector<vector<mint>> g(1 << n, vector<mint>(n));
    for (int mask = 0; mask < (1 << n); ++mask) {
      for (int i = 0; i < n; ++i) {
        g[mask][i] = 1;
        for (int j = 0; j < n; ++j) {
          if (mask & (1 << j)) {
            g[mask][i] *= a[i][j];
          }
        }
      }
    }
    vector<mint> f(1 << n);
    for (int mask = 1; mask < (1 << n); ++mask) {
      f[mask] = 1;
      for (int sub_mask = mask; sub_mask > 0;
           sub_mask = ((sub_mask - 1) & mask)) {
        if (sub_mask < mask) {
          mint val = f[sub_mask];
          int other_mask = mask - sub_mask;
          for (int i = 0; i < n; ++i) {
            if (sub_mask & (1 << i)) {
              val *= g[other_mask][i];
            }
          }
          f[mask] -= val;
        }
      }
    }
    mint result = 0;
    for (int mask = 1; mask < (1 << n); ++mask) {
      int count = 0;
      for (int i = 0; i < n; ++i) {
        if (mask & (1 << i)) {
          ++count;
        }
      }
      mint val = f[mask] * mint(count);
      int other_mask = (1 << n) - 1 - mask;
      for (int i = 0; i < n; ++i) {
        if (mask & (1 << i)) {
          val *= g[other_mask][i];
        }
      }
      result += val;
    }
    printf("%d\n", result.v);
  }
  return 0;
}

### 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;
int inv(int v) {
  return v <= 1 ? v
                : (1000000007) - (long long)inv((1000000007) % v) *
                                     ((1000000007) / v) % (1000000007);
}
struct mint {
  int v;
  mint() = default;
  mint(long long v) : v((v % (1000000007) + (1000000007)) % (1000000007)) {}
  mint& operator+=(mint a) {
    v = (v + a.v) % (1000000007);
    return *this;
  }
  mint& operator-=(mint a) {
    v = (v - a.v + (1000000007)) % (1000000007);
    return *this;
  }
  mint& operator*=(mint a) {
    v = ((long long)v * a.v) % (1000000007);
    return *this;
  }
  mint& operator/=(mint a) {
    v = ((long long)v * inv(a.v)) % (1000000007);
    return *this;
  }
  friend mint operator+(mint a, mint b) { return mint(a.v + b.v); }
  friend mint operator-(mint a, mint b) { return mint(a.v - b.v); }
  friend mint operator*(mint a, mint b) { return mint((long long)a.v * b.v); }
  friend mint operator/(mint a, mint b) {
    return mint((long long)a.v * inv(b.v));
  }
};
mint pow(mint base, long long exponent) {
  if (exponent <= 0) {
    return 1;
  }
  mint ret = 1;
  while (exponent) {
    if (exponent & 1) {
      ret *= base;
    }
    exponent >>= 1;
    base = base * base;
  }
  return ret;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  while (cin >> n) {
    vector<int> w(n);
    for (int i = 0; i < n; ++i) {
      cin >> w[i];
    }
    vector<vector<mint>> a(n, vector<mint>(n));
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        a[i][j] = mint(w[i]) / mint(w[i] + w[j]);
      }
    }
    vector<vector<mint>> g(1 << n, vector<mint>(n));
    for (int mask = 0; mask < (1 << n); ++mask) {
      for (int i = 0; i < n; ++i) {
        g[mask][i] = 1;
        for (int j = 0; j < n; ++j) {
          if (mask & (1 << j)) {
            g[mask][i] *= a[i][j];
          }
        }
      }
    }
    vector<mint> f(1 << n);
    for (int mask = 1; mask < (1 << n); ++mask) {
      f[mask] = 1;
      for (int sub_mask = mask; sub_mask > 0;
           sub_mask = ((sub_mask - 1) & mask)) {
        if (sub_mask < mask) {
          mint val = f[sub_mask];
          int other_mask = mask - sub_mask;
          for (int i = 0; i < n; ++i) {
            if (sub_mask & (1 << i)) {
              val *= g[other_mask][i];
            }
          }
          f[mask] -= val;
        }
      }
    }
    mint result = 0;
    for (int mask = 1; mask < (1 << n); ++mask) {
      int count = 0;
      for (int i = 0; i < n; ++i) {
        if (mask & (1 << i)) {
          ++count;
        }
      }
      mint val = f[mask] * mint(count);
      int other_mask = (1 << n) - 1 - mask;
      for (int i = 0; i < n; ++i) {
        if (mask & (1 << i)) {
          val *= g[other_mask][i];
        }
      }
      result += val;
    }
    printf("%d\n", result.v);
  }
  return 0;
}

```