#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  char c = getchar();
  long long su = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
  return su * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
long long P[15][15], f[1 << 15], ans, n, a[15];
long long quickPower(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = res * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return res;
}
long long work(long long x, long long y) {
  vector<long long> g;
  long long res = 1;
  for (long long i = 1; i <= n; i++) {
    if ((y >> (i - 1)) % 2) g.push_back(i);
  }
  for (long long i = 1; i <= n; i++)
    if ((x >> (i - 1)) % 2) {
      long long now = 1;
      for (auto u : g) {
        now = now * P[u][i] % 1000000007;
      }
      res = res * now % 1000000007;
    }
  return res;
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++)
      P[i][j] = a[i] * quickPower(a[i] + a[j], 1000000007 - 2) % 1000000007;
  for (long long stat = 1; stat < (1 << n); stat++) {
    f[stat] = 1;
    for (long long S = (stat - 1) & stat; S; S = (S - 1) & stat) {
      f[stat] = (f[stat] - f[S] * work(stat ^ S, S) % 1000000007 + 1000000007) %
                1000000007;
      if (S == 0) break;
    }
    ans = (ans + __builtin_popcount(stat) * f[stat] % 1000000007 *
                     work(((1 << n) - 1) ^ stat, stat)) %
          1000000007;
  }
  writeln(ans);
}

### 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>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  char c = getchar();
  long long su = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
  return su * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
long long P[15][15], f[1 << 15], ans, n, a[15];
long long quickPower(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = res * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return res;
}
long long work(long long x, long long y) {
  vector<long long> g;
  long long res = 1;
  for (long long i = 1; i <= n; i++) {
    if ((y >> (i - 1)) % 2) g.push_back(i);
  }
  for (long long i = 1; i <= n; i++)
    if ((x >> (i - 1)) % 2) {
      long long now = 1;
      for (auto u : g) {
        now = now * P[u][i] % 1000000007;
      }
      res = res * now % 1000000007;
    }
  return res;
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++)
      P[i][j] = a[i] * quickPower(a[i] + a[j], 1000000007 - 2) % 1000000007;
  for (long long stat = 1; stat < (1 << n); stat++) {
    f[stat] = 1;
    for (long long S = (stat - 1) & stat; S; S = (S - 1) & stat) {
      f[stat] = (f[stat] - f[S] * work(stat ^ S, S) % 1000000007 + 1000000007) %
                1000000007;
      if (S == 0) break;
    }
    ans = (ans + __builtin_popcount(stat) * f[stat] % 1000000007 *
                     work(((1 << n) - 1) ^ stat, stat)) %
          1000000007;
  }
  writeln(ans);
}

```