#include <bits/stdc++.h>
template <typename T>
inline void rd(T& x) {
  int si = 1;
  char c = getchar();
  x = 0;
  while (!isdigit(c)) si = c == '-' ? -1 : si, c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  x *= si;
}
template <typename T, typename... Args>
inline void rd(T& x, Args&... args) {
  rd(x);
  rd(args...);
}
const int N = 15, M = (1 << 14) + 5, Inf = 0x3f3f3f3f, Mod = 1e9 + 7;
const long long InfLL = 0x3f3f3f3f3f3f3f3fLL;
int QPow(int a, int b) {
  int ret = 1, bas = a;
  for (; b; b >>= 1, bas = 1LL * bas * bas % Mod)
    if (b & 1) ret = 1LL * ret * bas % Mod;
  return ret;
}
int n, a[N], g[N][N], w[N][M], f[M][N];
int main() {
  int test_case_cnt = 1;
  while (test_case_cnt--) {
    rd(n);
    for (int i = 0; i < n; ++i) rd(a[i]);
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        g[i][j] = 1LL * a[i] * QPow(a[i] + a[j], Mod - 2) % Mod;
    for (int i = 0; i < n; ++i)
      for (int S = 0; S < 1 << n; ++S) {
        w[i][S] = 1;
        for (int j = 0; j < n; ++j)
          if ((S >> j) & 1) w[i][S] = 1LL * w[i][S] * g[i][j] % Mod;
      }
    int ans = 0;
    for (int S = 1; S < 1 << n; ++S) {
      if (S & -S == S) {
        f[S][1] = 1;
        continue;
      }
      for (int T = (S - 1) & S; T; T = (T - 1) & S) {
        int p = 1;
        for (int i = 0; i < n; ++i)
          if ((T >> i) & 1) p = 1LL * p * w[i][S ^ T] % Mod;
        for (int i = 2; i <= n; ++i) {
          f[S][i] = (f[S][i] + 1LL * f[T][1] * f[S ^ T][i - 1] % Mod * p) % Mod;
          if (S == (1 << n) - 1) {
            int siz = __builtin_popcount(T);
            ans =
                (ans + 1LL * f[T][1] * f[S ^ T][i - 1] % Mod * p % Mod * siz) %
                Mod;
          }
        }
      }
      int tot = 0;
      for (int i = 2; i <= n; ++i) tot = (tot + f[S][i]) % Mod;
      f[S][1] = (1 - tot + Mod) % Mod;
    }
    ans = (ans + 1LL * n * f[(1 << n) - 1][1]) % Mod;
    printf("%d\n", ans);
  }
  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>
template <typename T>
inline void rd(T& x) {
  int si = 1;
  char c = getchar();
  x = 0;
  while (!isdigit(c)) si = c == '-' ? -1 : si, c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  x *= si;
}
template <typename T, typename... Args>
inline void rd(T& x, Args&... args) {
  rd(x);
  rd(args...);
}
const int N = 15, M = (1 << 14) + 5, Inf = 0x3f3f3f3f, Mod = 1e9 + 7;
const long long InfLL = 0x3f3f3f3f3f3f3f3fLL;
int QPow(int a, int b) {
  int ret = 1, bas = a;
  for (; b; b >>= 1, bas = 1LL * bas * bas % Mod)
    if (b & 1) ret = 1LL * ret * bas % Mod;
  return ret;
}
int n, a[N], g[N][N], w[N][M], f[M][N];
int main() {
  int test_case_cnt = 1;
  while (test_case_cnt--) {
    rd(n);
    for (int i = 0; i < n; ++i) rd(a[i]);
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        g[i][j] = 1LL * a[i] * QPow(a[i] + a[j], Mod - 2) % Mod;
    for (int i = 0; i < n; ++i)
      for (int S = 0; S < 1 << n; ++S) {
        w[i][S] = 1;
        for (int j = 0; j < n; ++j)
          if ((S >> j) & 1) w[i][S] = 1LL * w[i][S] * g[i][j] % Mod;
      }
    int ans = 0;
    for (int S = 1; S < 1 << n; ++S) {
      if (S & -S == S) {
        f[S][1] = 1;
        continue;
      }
      for (int T = (S - 1) & S; T; T = (T - 1) & S) {
        int p = 1;
        for (int i = 0; i < n; ++i)
          if ((T >> i) & 1) p = 1LL * p * w[i][S ^ T] % Mod;
        for (int i = 2; i <= n; ++i) {
          f[S][i] = (f[S][i] + 1LL * f[T][1] * f[S ^ T][i - 1] % Mod * p) % Mod;
          if (S == (1 << n) - 1) {
            int siz = __builtin_popcount(T);
            ans =
                (ans + 1LL * f[T][1] * f[S ^ T][i - 1] % Mod * p % Mod * siz) %
                Mod;
          }
        }
      }
      int tot = 0;
      for (int i = 2; i <= n; ++i) tot = (tot + f[S][i]) % Mod;
      f[S][1] = (1 - tot + Mod) % Mod;
    }
    ans = (ans + 1LL * n * f[(1 << n) - 1][1]) % Mod;
    printf("%d\n", ans);
  }
  return 0;
}

```