#include <bits/stdc++.h>
using namespace std;
const long long mod = static_cast<long long>(1e9 + 7);
long long P, Q = 1;
int n;
long long a[15];
pair<long long, long long> f[15][1 << 14];
pair<long long, long long> g[5000000];
pair<long long, long long> p[1 << 14];
pair<long long, long long> h[1 << 14];
int count(int x) {
  if (!x) return 0;
  return (x & 1) + count(x >> 1);
}
void Init_F() {
  for (int i = 1; i <= n; ++i) {
    for (int s = 0; s < (1 << n); ++s) {
      if ((1 << (i - 1)) & s) {
        continue;
      }
      f[i][s] = make_pair(1, 1);
      for (int j = 1; j <= n; ++j) {
        if ((1 << (j - 1)) & s) {
          f[i][s].first = f[i][s].first * a[i] % mod;
          f[i][s].second = f[i][s].second * (a[i] + a[j]) % mod;
        }
      }
    }
  }
}
int GetIndexBase3(int left, int right) {
  int idx = 0;
  for (int i = n; i > 0; --i) {
    idx *= 3;
    if (left & (1 << (i - 1))) {
      idx += 1;
    } else if (right & (1 << (i - 1))) {
      idx += 2;
    }
  }
  return idx;
}
void Recursively_Init_G(int left, int right, int cur) {
  if (cur > n) {
    int idx = GetIndexBase3(left, right);
    g[idx] = make_pair(1, 1);
    for (int i = 1; i <= n; ++i) {
      if ((1 << (i - 1)) & left) {
        g[idx].first = g[idx].first * f[i][right].first % mod;
        g[idx].second = g[idx].second * f[i][right].second % mod;
      }
    }
  } else {
    Recursively_Init_G(left, right, cur + 1);
    Recursively_Init_G(left + (1 << (cur - 1)), right, cur + 1);
    Recursively_Init_G(left, right + (1 << (cur - 1)), cur + 1);
  }
}
void Init_P() {
  for (int i = 1; i < (1 << n); ++i) {
    p[i] = make_pair(1, 1);
    for (int j = 1; j < i; ++j) {
      if ((i | j) == i) {
        long long upper = p[j].first * g[GetIndexBase3(j, i - j)].first % mod;
        long long lower = p[j].second * g[GetIndexBase3(j, i - j)].second % mod;
        p[i].first =
            ((p[i].first * lower - upper * p[i].second) % mod + mod) % mod;
        p[i].second = p[i].second * lower % mod;
      }
    }
  }
}
void Solve() {
  Init_F();
  Recursively_Init_G(0, 0, 1);
  Init_P();
  for (int i = 1; i < (1 << n); ++i) {
    int j = (1 << n) - 1 - i;
    long long upper =
        count(i) * p[i].first % mod * g[GetIndexBase3(i, j)].first % mod;
    long long lower = p[i].second * g[GetIndexBase3(i, j)].second % mod;
    P = (P * lower + Q * upper) % mod;
    Q = Q * lower % mod;
  }
}
long long Pow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = res * x % mod;
    }
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  Solve();
  cout << P * Pow(Q, mod - 2) % mod << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

<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;
const long long mod = static_cast<long long>(1e9 + 7);
long long P, Q = 1;
int n;
long long a[15];
pair<long long, long long> f[15][1 << 14];
pair<long long, long long> g[5000000];
pair<long long, long long> p[1 << 14];
pair<long long, long long> h[1 << 14];
int count(int x) {
  if (!x) return 0;
  return (x & 1) + count(x >> 1);
}
void Init_F() {
  for (int i = 1; i <= n; ++i) {
    for (int s = 0; s < (1 << n); ++s) {
      if ((1 << (i - 1)) & s) {
        continue;
      }
      f[i][s] = make_pair(1, 1);
      for (int j = 1; j <= n; ++j) {
        if ((1 << (j - 1)) & s) {
          f[i][s].first = f[i][s].first * a[i] % mod;
          f[i][s].second = f[i][s].second * (a[i] + a[j]) % mod;
        }
      }
    }
  }
}
int GetIndexBase3(int left, int right) {
  int idx = 0;
  for (int i = n; i > 0; --i) {
    idx *= 3;
    if (left & (1 << (i - 1))) {
      idx += 1;
    } else if (right & (1 << (i - 1))) {
      idx += 2;
    }
  }
  return idx;
}
void Recursively_Init_G(int left, int right, int cur) {
  if (cur > n) {
    int idx = GetIndexBase3(left, right);
    g[idx] = make_pair(1, 1);
    for (int i = 1; i <= n; ++i) {
      if ((1 << (i - 1)) & left) {
        g[idx].first = g[idx].first * f[i][right].first % mod;
        g[idx].second = g[idx].second * f[i][right].second % mod;
      }
    }
  } else {
    Recursively_Init_G(left, right, cur + 1);
    Recursively_Init_G(left + (1 << (cur - 1)), right, cur + 1);
    Recursively_Init_G(left, right + (1 << (cur - 1)), cur + 1);
  }
}
void Init_P() {
  for (int i = 1; i < (1 << n); ++i) {
    p[i] = make_pair(1, 1);
    for (int j = 1; j < i; ++j) {
      if ((i | j) == i) {
        long long upper = p[j].first * g[GetIndexBase3(j, i - j)].first % mod;
        long long lower = p[j].second * g[GetIndexBase3(j, i - j)].second % mod;
        p[i].first =
            ((p[i].first * lower - upper * p[i].second) % mod + mod) % mod;
        p[i].second = p[i].second * lower % mod;
      }
    }
  }
}
void Solve() {
  Init_F();
  Recursively_Init_G(0, 0, 1);
  Init_P();
  for (int i = 1; i < (1 << n); ++i) {
    int j = (1 << n) - 1 - i;
    long long upper =
        count(i) * p[i].first % mod * g[GetIndexBase3(i, j)].first % mod;
    long long lower = p[i].second * g[GetIndexBase3(i, j)].second % mod;
    P = (P * lower + Q * upper) % mod;
    Q = Q * lower % mod;
  }
}
long long Pow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = res * x % mod;
    }
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  Solve();
  cout << P * Pow(Q, mod - 2) % mod << endl;
  return 0;
}

```