#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int mod = inf + 7;
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline void _inc(int &x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline void _dec(int &x, int y) {
  if ((x -= y) < 0) {
    x += mod;
  }
}
inline int _binpow(int x, int pw) {
  int res = 1;
  int tmp = x;
  while (pw > 0) {
    if (pw & 1) {
      res = _mul(res, tmp);
    }
    tmp = _mul(tmp, tmp);
    pw >>= 1;
  }
  return res;
}
inline int _div(int a, int b) { return _mul(a, _binpow(b, mod - 2)); }
int slow(int n, vector<int> a) {
  vector<vector<int>> g(n, vector<int>(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      g[i][j] = _div(a[i], a[i] + a[j]);
    }
  }
  vector<pair<int, int>> edges;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) edges.emplace_back(i, j);
  }
  int m = (int)edges.size();
  int res = 0;
  for (int mask = 0; mask < (1 << m); mask++) {
    int prob = 1;
    vector<vector<int>> f(n, vector<int>(n));
    for (int i = 0; i < m; i++) {
      auto [u, v] = edges[i];
      int eprob = 0;
      if ((1 << i) & mask) {
        f[u][v] = 1;
        eprob = g[u][v];
      } else {
        f[v][u] = 1;
        eprob = g[v][u];
      }
      prob = _mul(prob, eprob);
    }
    for (int k = 0; k < n; k++) {
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
          f[i][j] |= f[i][k] & f[k][j];
        }
      }
    }
    for (int v = 0; v < n; v++) f[v][v] = 1;
    int sz = 0;
    for (int v = 0; v < n; v++) {
      int ok = 1;
      for (int u = 0; u < n; u++) ok &= f[v][u];
      sz += ok;
    }
    _inc(res, _mul(sz, prob));
  }
  return res;
}
int fast(int n, vector<int> a) {
  vector<vector<int>> g(n, vector<int>(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      g[i][j] = _div(a[i], a[i] + a[j]);
    }
  }
  vector<int> comp2(1 << n);
  for (int mask = 0; mask < (1 << n); mask++) {
    vector<int> in;
    for (int i = 0; i < n; i++) {
      if ((1 << i) & mask) in.push_back(i);
    }
    for (int A = mask; A > 0; A = (A - 1) & mask) {
      int B = mask ^ A;
      if (!A || !B) continue;
      int prob = 1;
      for (int u : in) {
        if ((1 << u) & B) continue;
        for (int v : in) {
          if ((1 << v) & A) continue;
          prob = _mul(prob, g[u][v]);
        }
      }
      _inc(comp2[mask], _mul(prob, comp2[A]));
    }
    comp2[mask] = _sub(1, comp2[mask]);
  }
  int res = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    vector<int> in, out;
    for (int i = 0; i < n; i++) {
      if ((1 << i) & mask) {
        in.push_back(i);
      } else {
        out.push_back(i);
      }
    }
    int coef = 1;
    for (int u : in) {
      for (int v : out) {
        coef = _mul(coef, g[u][v]);
      }
    }
    int prob = _mul(coef, comp2[mask]);
    _inc(res, _mul(prob, (int)in.size()));
  }
  return res;
}
void work() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int res = fast(n, a);
  cout << res << "\n";
}
void test() {
  mt19937 rnd(42);
  const int MAXN = 5;
  const int MAXX = 10;
  while (1) {
    int n = rnd() % MAXN + 1;
    vector<int> a(n);
    for (int i = 0; i < n; i++) a[i] = rnd() % MAXX + 1;
    int fs = fast(n, a);
    int sl = slow(n, a);
    if (fs == sl) {
      cout << "OK(" << fs << ")" << endl;
    } else {
      cout << "WA\n";
      cout << "exp = " << sl << "\n";
      cout << "fnd = " << fs << "\n";
      cout << n << "\n";
      for (int i = 0; i < n; i++) cout << a[i] << " ";
      cout << "\n";
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  work();
  return 0;
}

### Prompt

Please formulate 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;
const int inf = 1e9;
const long long inf64 = 1e18;
const int mod = inf + 7;
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline void _inc(int &x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline void _dec(int &x, int y) {
  if ((x -= y) < 0) {
    x += mod;
  }
}
inline int _binpow(int x, int pw) {
  int res = 1;
  int tmp = x;
  while (pw > 0) {
    if (pw & 1) {
      res = _mul(res, tmp);
    }
    tmp = _mul(tmp, tmp);
    pw >>= 1;
  }
  return res;
}
inline int _div(int a, int b) { return _mul(a, _binpow(b, mod - 2)); }
int slow(int n, vector<int> a) {
  vector<vector<int>> g(n, vector<int>(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      g[i][j] = _div(a[i], a[i] + a[j]);
    }
  }
  vector<pair<int, int>> edges;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) edges.emplace_back(i, j);
  }
  int m = (int)edges.size();
  int res = 0;
  for (int mask = 0; mask < (1 << m); mask++) {
    int prob = 1;
    vector<vector<int>> f(n, vector<int>(n));
    for (int i = 0; i < m; i++) {
      auto [u, v] = edges[i];
      int eprob = 0;
      if ((1 << i) & mask) {
        f[u][v] = 1;
        eprob = g[u][v];
      } else {
        f[v][u] = 1;
        eprob = g[v][u];
      }
      prob = _mul(prob, eprob);
    }
    for (int k = 0; k < n; k++) {
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
          f[i][j] |= f[i][k] & f[k][j];
        }
      }
    }
    for (int v = 0; v < n; v++) f[v][v] = 1;
    int sz = 0;
    for (int v = 0; v < n; v++) {
      int ok = 1;
      for (int u = 0; u < n; u++) ok &= f[v][u];
      sz += ok;
    }
    _inc(res, _mul(sz, prob));
  }
  return res;
}
int fast(int n, vector<int> a) {
  vector<vector<int>> g(n, vector<int>(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      g[i][j] = _div(a[i], a[i] + a[j]);
    }
  }
  vector<int> comp2(1 << n);
  for (int mask = 0; mask < (1 << n); mask++) {
    vector<int> in;
    for (int i = 0; i < n; i++) {
      if ((1 << i) & mask) in.push_back(i);
    }
    for (int A = mask; A > 0; A = (A - 1) & mask) {
      int B = mask ^ A;
      if (!A || !B) continue;
      int prob = 1;
      for (int u : in) {
        if ((1 << u) & B) continue;
        for (int v : in) {
          if ((1 << v) & A) continue;
          prob = _mul(prob, g[u][v]);
        }
      }
      _inc(comp2[mask], _mul(prob, comp2[A]));
    }
    comp2[mask] = _sub(1, comp2[mask]);
  }
  int res = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    vector<int> in, out;
    for (int i = 0; i < n; i++) {
      if ((1 << i) & mask) {
        in.push_back(i);
      } else {
        out.push_back(i);
      }
    }
    int coef = 1;
    for (int u : in) {
      for (int v : out) {
        coef = _mul(coef, g[u][v]);
      }
    }
    int prob = _mul(coef, comp2[mask]);
    _inc(res, _mul(prob, (int)in.size()));
  }
  return res;
}
void work() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int res = fast(n, a);
  cout << res << "\n";
}
void test() {
  mt19937 rnd(42);
  const int MAXN = 5;
  const int MAXX = 10;
  while (1) {
    int n = rnd() % MAXN + 1;
    vector<int> a(n);
    for (int i = 0; i < n; i++) a[i] = rnd() % MAXX + 1;
    int fs = fast(n, a);
    int sl = slow(n, a);
    if (fs == sl) {
      cout << "OK(" << fs << ")" << endl;
    } else {
      cout << "WA\n";
      cout << "exp = " << sl << "\n";
      cout << "fnd = " << fs << "\n";
      cout << n << "\n";
      for (int i = 0; i < n; i++) cout << a[i] << " ";
      cout << "\n";
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  work();
  return 0;
}

```