#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int r = 0, f = 1;
  char c = getchar();
  while (c < 48 || c > 57) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) r = r * 10 + (c & 15), c = getchar();
  return r * f;
}
inline void write(int x) {
  char c[20], len = 0;
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) x = -x, putchar('-');
  while (x) c[len++] = x % 10, x /= 10;
  while (len) putchar(c[--len] + 48);
}
const int mod = 1e9 + 7;
struct M {
  int x;
  inline M(int a = 0) : x(a) {}
  inline M operator+(const M& p) const {
    return x + p.x >= mod ? x + p.x - mod : x + p.x;
  }
  inline M operator-() const { return x ? mod - x : 0; }
  inline M operator-(const M& p) const {
    return x - p.x < 0 ? x - p.x + mod : x - p.x;
  }
  inline M operator*(const M& p) const { return (long long)x * p.x % mod; }
  inline bool operator==(const int& p) const { return x == p; }
  inline void operator+=(const M& p) { *this = *this + p; }
  inline void operator-=(const M& p) { *this = *this - p; }
  inline void operator*=(const M& p) { *this = *this * p; }
};
inline void write(const M& x) { write(x.x); }
inline M qpow(M x, int y) {
  M res = 1;
  for (; y; y >>= 1) y & 1 ? (res = res * x) : 1, x *= x;
  return res;
}
inline M inv(M x) { return qpow(x, mod - 2); }
const int N = 14, L = 1 << 14;
int n, lg[L];
M a[N], p[N][N], f[N][L], h[N][L];
inline M G(int S, int T) {
  M res = 1;
  while (S) res *= h[lg[S & -S]][T], S ^= S & -S;
  return res;
}
int main() {
  n = read();
  for (int i = 0; i < n; ++i) a[i] = read();
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i ^ j) p[i][j] = a[i] * inv(a[i] + a[j]);
  int all = (1 << n) - 1;
  for (int i = 2; i <= all; ++i) lg[i] = lg[i >> 1] + 1;
  for (int i = 0; i < n; ++i) {
    h[i][0] = 1;
    for (int S = 1; S <= all; ++S)
      if (S >> i & 1 ^ 1) h[i][S] = h[i][S ^ (S & -S)] * p[lg[S & -S]][i];
  }
  for (int i = 0; i < n; ++i)
    for (int S = 1; S <= all; ++S)
      if (S >> i & 1) {
        f[i][S] = 1;
        for (int T = (S - 1) & S; T; T = (T - 1) & S)
          if (T >> i & 1) f[i][S] -= f[i][T] * G(T, S ^ T);
      }
  M ans = 0;
  for (int i = 0; i < n; ++i) ans += f[i][all];
  write(ans), putchar('\n');
}

### 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;
inline int read() {
  int r = 0, f = 1;
  char c = getchar();
  while (c < 48 || c > 57) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) r = r * 10 + (c & 15), c = getchar();
  return r * f;
}
inline void write(int x) {
  char c[20], len = 0;
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) x = -x, putchar('-');
  while (x) c[len++] = x % 10, x /= 10;
  while (len) putchar(c[--len] + 48);
}
const int mod = 1e9 + 7;
struct M {
  int x;
  inline M(int a = 0) : x(a) {}
  inline M operator+(const M& p) const {
    return x + p.x >= mod ? x + p.x - mod : x + p.x;
  }
  inline M operator-() const { return x ? mod - x : 0; }
  inline M operator-(const M& p) const {
    return x - p.x < 0 ? x - p.x + mod : x - p.x;
  }
  inline M operator*(const M& p) const { return (long long)x * p.x % mod; }
  inline bool operator==(const int& p) const { return x == p; }
  inline void operator+=(const M& p) { *this = *this + p; }
  inline void operator-=(const M& p) { *this = *this - p; }
  inline void operator*=(const M& p) { *this = *this * p; }
};
inline void write(const M& x) { write(x.x); }
inline M qpow(M x, int y) {
  M res = 1;
  for (; y; y >>= 1) y & 1 ? (res = res * x) : 1, x *= x;
  return res;
}
inline M inv(M x) { return qpow(x, mod - 2); }
const int N = 14, L = 1 << 14;
int n, lg[L];
M a[N], p[N][N], f[N][L], h[N][L];
inline M G(int S, int T) {
  M res = 1;
  while (S) res *= h[lg[S & -S]][T], S ^= S & -S;
  return res;
}
int main() {
  n = read();
  for (int i = 0; i < n; ++i) a[i] = read();
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i ^ j) p[i][j] = a[i] * inv(a[i] + a[j]);
  int all = (1 << n) - 1;
  for (int i = 2; i <= all; ++i) lg[i] = lg[i >> 1] + 1;
  for (int i = 0; i < n; ++i) {
    h[i][0] = 1;
    for (int S = 1; S <= all; ++S)
      if (S >> i & 1 ^ 1) h[i][S] = h[i][S ^ (S & -S)] * p[lg[S & -S]][i];
  }
  for (int i = 0; i < n; ++i)
    for (int S = 1; S <= all; ++S)
      if (S >> i & 1) {
        f[i][S] = 1;
        for (int T = (S - 1) & S; T; T = (T - 1) & S)
          if (T >> i & 1) f[i][S] -= f[i][T] * G(T, S ^ T);
      }
  M ans = 0;
  for (int i = 0; i < n; ++i) ans += f[i][all];
  write(ans), putchar('\n');
}

```