#include <bits/stdc++.h>
using namespace std;
inline int min(const int a, const int b) { return (a <= b) ? a : b; }
inline int min3(const int a, const int b, const int c) {
  return (a <= b && a <= c) ? a : ((b <= c) ? b : c);
}
inline int max(const int a, const int b) { return (a >= b) ? a : b; }
inline int max3(const int a, const int b, const int c) {
  return (a >= b && a >= c) ? a : ((b >= c) ? b : c);
}
inline int gcd(const int a, const int b) { return b == 0 ? a : gcd(b, a % b); }
inline int lcm(const int a, const int b) { return 1LL * a * b / gcd(a, b); }
inline int pow(int x, int t, int mod) {
  int g = 1;
  while (t) {
    if (t & 1) g = 1LL * g * x % mod;
    x = 1LL * x * x % mod, t >>= 1;
  }
  return g;
}
inline int bits(int x) {
  x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
  x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
  x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
  x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
  x = (x & 0x0000FFFF) + ((x >> 16) & 0x0000FFFF);
  return x;
}
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) f |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
inline void init_inverse(int *inv, int num, int mod) {
  inv[1] = 1;
  for (int i = 2; i <= num; i++) {
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  }
}
int inv[2000009];
int n, a[19];
int g[4782969], b[4782969], d[4782969], c[15];
int f[16384];
inline int pow(int a, int b) { return pow(a, b, 1000000007); }
int main() {
  init_inverse(inv, 2000000, 1000000007);
  n = read();
  for (int i = 0; i < n; i++) a[i] = read();
  int mxo = pow(3, n);
  g[0] = 1;
  for (int o = 1; o <= mxo - 1; o++) {
    c[0]++;
    for (int k = 0; k < n; k++)
      if (c[k] == 3)
        c[k] = 0, c[k + 1]++;
      else if (c[k]) {
        g[o] = g[o - c[k] * pow(3, k)];
        b[o] |= 1 << k;
        if (c[k] == 2) d[o] |= 1 << k;
        for (int i = k + 1; i <= n - 1; i++)
          if (c[i]) {
            if (c[k] != c[i]) {
              if (c[k] == 2) {
                g[o] = 1LL * g[o] * inv[a[k] + a[i]] % 1000000007 * a[k] %
                       1000000007;
              } else {
                g[o] = 1LL * g[o] * inv[a[k] + a[i]] % 1000000007 * a[i] %
                       1000000007;
              }
            }
            b[o] |= 1 << i;
            if (c[i] == 2) d[o] |= 1 << i;
          }
        break;
      }
  }
  for (int o = 0; o < 1 << n; o++) f[o] = 1;
  for (int o = 0; o < mxo; o++)
    if (d[o] && d[o] != b[o])
      f[b[o]] = (f[b[o]] + 1000000007 - 1LL * g[o] * f[d[o]] % 1000000007) %
                1000000007;
  int ans = 0;
  for (int o = 0; o < mxo; o++)
    if (b[o] + 1 == (1 << n))
      ans =
          (ans + 1LL * f[d[o]] * g[o] % 1000000007 * bits(d[o]) % 1000000007) %
          1000000007;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
inline int min(const int a, const int b) { return (a <= b) ? a : b; }
inline int min3(const int a, const int b, const int c) {
  return (a <= b && a <= c) ? a : ((b <= c) ? b : c);
}
inline int max(const int a, const int b) { return (a >= b) ? a : b; }
inline int max3(const int a, const int b, const int c) {
  return (a >= b && a >= c) ? a : ((b >= c) ? b : c);
}
inline int gcd(const int a, const int b) { return b == 0 ? a : gcd(b, a % b); }
inline int lcm(const int a, const int b) { return 1LL * a * b / gcd(a, b); }
inline int pow(int x, int t, int mod) {
  int g = 1;
  while (t) {
    if (t & 1) g = 1LL * g * x % mod;
    x = 1LL * x * x % mod, t >>= 1;
  }
  return g;
}
inline int bits(int x) {
  x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
  x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
  x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
  x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
  x = (x & 0x0000FFFF) + ((x >> 16) & 0x0000FFFF);
  return x;
}
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) f |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
inline void init_inverse(int *inv, int num, int mod) {
  inv[1] = 1;
  for (int i = 2; i <= num; i++) {
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  }
}
int inv[2000009];
int n, a[19];
int g[4782969], b[4782969], d[4782969], c[15];
int f[16384];
inline int pow(int a, int b) { return pow(a, b, 1000000007); }
int main() {
  init_inverse(inv, 2000000, 1000000007);
  n = read();
  for (int i = 0; i < n; i++) a[i] = read();
  int mxo = pow(3, n);
  g[0] = 1;
  for (int o = 1; o <= mxo - 1; o++) {
    c[0]++;
    for (int k = 0; k < n; k++)
      if (c[k] == 3)
        c[k] = 0, c[k + 1]++;
      else if (c[k]) {
        g[o] = g[o - c[k] * pow(3, k)];
        b[o] |= 1 << k;
        if (c[k] == 2) d[o] |= 1 << k;
        for (int i = k + 1; i <= n - 1; i++)
          if (c[i]) {
            if (c[k] != c[i]) {
              if (c[k] == 2) {
                g[o] = 1LL * g[o] * inv[a[k] + a[i]] % 1000000007 * a[k] %
                       1000000007;
              } else {
                g[o] = 1LL * g[o] * inv[a[k] + a[i]] % 1000000007 * a[i] %
                       1000000007;
              }
            }
            b[o] |= 1 << i;
            if (c[i] == 2) d[o] |= 1 << i;
          }
        break;
      }
  }
  for (int o = 0; o < 1 << n; o++) f[o] = 1;
  for (int o = 0; o < mxo; o++)
    if (d[o] && d[o] != b[o])
      f[b[o]] = (f[b[o]] + 1000000007 - 1LL * g[o] * f[d[o]] % 1000000007) %
                1000000007;
  int ans = 0;
  for (int o = 0; o < mxo; o++)
    if (b[o] + 1 == (1 << n))
      ans =
          (ans + 1LL * f[d[o]] * g[o] % 1000000007 * bits(d[o]) % 1000000007) %
          1000000007;
  printf("%d\n", ans);
  return 0;
}

```