#include <bits/stdc++.h>
using namespace std;
namespace IO {
char in[1 << 24];
char const *o;
void init_in() {
  o = in;
  in[fread(in, 1, sizeof(in) - 4, stdin)] = 0;
}
int rd() {
  unsigned u = 0, s = 0;
  while (*o && *o <= 32) ++o;
  if (*o == '-')
    s = ~s, ++o;
  else if (*o == '+')
    ++o;
  while (*o >= '0' && *o <= '9') u = (u << 3) + (u << 1) + (*o++ - '0');
  return static_cast<int>((u ^ s) + !!s);
}
char *rdstr(char *s) {
  while (*o && *o <= 32) ++o;
  while (*o > 32) *s++ = *o++;
  return s;
}
}  // namespace IO
template <typename T>
void chkmax(T &a, const T &b) {
  a = max(a, b);
}
template <typename T>
void chkmin(T &a, const T &b) {
  a = min(a, b);
}
pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
using IO::rd;
const int N = 14, S = 1 << N;
int a[N];
int n;
const long long mod = 1e9 + 7;
long long b[N][N];
long long h[N][S], pr[S];
inline void reduce(long long &x) {
  if (x >= mod) x -= mod;
}
inline long long g(int s, int t) {
  long long ret = 1;
  for (int i = 0; i < n; ++i) {
    if (s & (1 << i)) ret = ret * h[i][t] % mod;
  }
  return ret;
}
int Q_pow(long long x, int y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ret;
}
inline int inv(int x) { return Q_pow(x, mod - 2); }
void solve() {
  n = rd();
  int m = 1 << n;
  for (int i = 0; i < n; ++i) a[i] = rd();
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < i; ++j) {
      long long x = inv(a[i] + a[j]);
      b[i][j] = a[i] * x % mod;
      b[j][i] = a[j] * x % mod;
    }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      h[i][j] = 1;
      for (int k = 0; k < n; ++k) {
        if (k != i && j & (1 << k)) h[i][j] = h[i][j] * b[i][k] % mod;
      }
    }
  int mask = m - 1;
  for (int i = 1; i < m; ++i) {
    pr[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      reduce(pr[i] += mod - (g(j, i ^ j) * pr[j] % mod));
    }
  }
  long long ans = 0;
  for (int i = 1; i < m; ++i) {
    ans += pr[i] * g(i, mask ^ i) % mod * __builtin_popcount(i) % mod;
  }
  cout << ans % mod << '\n';
}
int main() {
  IO::init_in();
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
namespace IO {
char in[1 << 24];
char const *o;
void init_in() {
  o = in;
  in[fread(in, 1, sizeof(in) - 4, stdin)] = 0;
}
int rd() {
  unsigned u = 0, s = 0;
  while (*o && *o <= 32) ++o;
  if (*o == '-')
    s = ~s, ++o;
  else if (*o == '+')
    ++o;
  while (*o >= '0' && *o <= '9') u = (u << 3) + (u << 1) + (*o++ - '0');
  return static_cast<int>((u ^ s) + !!s);
}
char *rdstr(char *s) {
  while (*o && *o <= 32) ++o;
  while (*o > 32) *s++ = *o++;
  return s;
}
}  // namespace IO
template <typename T>
void chkmax(T &a, const T &b) {
  a = max(a, b);
}
template <typename T>
void chkmin(T &a, const T &b) {
  a = min(a, b);
}
pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
using IO::rd;
const int N = 14, S = 1 << N;
int a[N];
int n;
const long long mod = 1e9 + 7;
long long b[N][N];
long long h[N][S], pr[S];
inline void reduce(long long &x) {
  if (x >= mod) x -= mod;
}
inline long long g(int s, int t) {
  long long ret = 1;
  for (int i = 0; i < n; ++i) {
    if (s & (1 << i)) ret = ret * h[i][t] % mod;
  }
  return ret;
}
int Q_pow(long long x, int y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ret;
}
inline int inv(int x) { return Q_pow(x, mod - 2); }
void solve() {
  n = rd();
  int m = 1 << n;
  for (int i = 0; i < n; ++i) a[i] = rd();
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < i; ++j) {
      long long x = inv(a[i] + a[j]);
      b[i][j] = a[i] * x % mod;
      b[j][i] = a[j] * x % mod;
    }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      h[i][j] = 1;
      for (int k = 0; k < n; ++k) {
        if (k != i && j & (1 << k)) h[i][j] = h[i][j] * b[i][k] % mod;
      }
    }
  int mask = m - 1;
  for (int i = 1; i < m; ++i) {
    pr[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      reduce(pr[i] += mod - (g(j, i ^ j) * pr[j] % mod));
    }
  }
  long long ans = 0;
  for (int i = 1; i < m; ++i) {
    ans += pr[i] * g(i, mask ^ i) % mod * __builtin_popcount(i) % mod;
  }
  cout << ans % mod << '\n';
}
int main() {
  IO::init_in();
  solve();
  return 0;
}

```