#include <bits/stdc++.h>
using namespace std;
const int MX = (1 << 15) + 2, md = 1000000007;
int n, m, i, j, k, msk, a[15], b[MX], pos[15];
long long rp, rq, zp, zq, curp, curq, cp[MX], cq[MX], p[15][15], q[15][15];
long long pw(long long a, int b) {
  if (b == 0) return 1LL;
  if (b & 1) return (pw(a, b - 1) * a) % md;
  long long x = pw(a, b / 2);
  return (x * x) % md;
}
void mul(long long& xp, long long& xq, long long yp, long long yq) {
  xp = (xp * yp) % md;
  xq = (xq * yq) % md;
}
void sub(long long& xp, long long& xq, long long yp, long long yq) {
  xp = (xp * yq + md - ((xq * yp) % md)) % md;
  xq = (xq * yq) % md;
}
void inc(long long& xp, long long& xq, long long yp, long long yq) {
  xp = (xp * yq + xq * yp) % md;
  xq = (xq * yq) % md;
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    for (j = 0; j < i; j++) {
      p[i][j] = a[i];
      p[j][i] = a[j];
      q[i][j] = q[j][i] = a[i] + a[j];
    }
  }
  rp = 0;
  rq = 1;
  for (i = 1; i < (1 << n); i++) {
    b[i] = b[i / 2] + (i & 1);
    cp[i] = zp = 1;
    cq[i] = zq = 1;
    for (m = j = 0; j < n; j++)
      if ((i >> j) & 1) {
        pos[m++] = j;
        for (k = 0; k < n; k++)
          if (((i >> k) & 1) == 0) mul(zp, zq, p[j][k], q[j][k]);
      }
    for (msk = (i & (i - 1)); msk > 0; msk = ((msk - 1) & i)) {
      curp = cp[msk];
      curq = cq[msk];
      for (j = 0; j < m; j++)
        if ((msk >> pos[j]) & 1)
          for (k = 0; k < m; k++)
            if (((msk >> pos[k]) & 1) == 0)
              mul(curp, curq, p[pos[j]][pos[k]], q[pos[j]][pos[k]]);
      sub(cp[i], cq[i], curp, curq);
    }
    inc(rp, rq, (((b[i] * cp[i]) % md) * zp) % md, (cq[i] * zq) % md);
  }
  printf("%lld\n", (rp * pw(rq, md - 2)) % md);
  return 0;
}

### 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;
const int MX = (1 << 15) + 2, md = 1000000007;
int n, m, i, j, k, msk, a[15], b[MX], pos[15];
long long rp, rq, zp, zq, curp, curq, cp[MX], cq[MX], p[15][15], q[15][15];
long long pw(long long a, int b) {
  if (b == 0) return 1LL;
  if (b & 1) return (pw(a, b - 1) * a) % md;
  long long x = pw(a, b / 2);
  return (x * x) % md;
}
void mul(long long& xp, long long& xq, long long yp, long long yq) {
  xp = (xp * yp) % md;
  xq = (xq * yq) % md;
}
void sub(long long& xp, long long& xq, long long yp, long long yq) {
  xp = (xp * yq + md - ((xq * yp) % md)) % md;
  xq = (xq * yq) % md;
}
void inc(long long& xp, long long& xq, long long yp, long long yq) {
  xp = (xp * yq + xq * yp) % md;
  xq = (xq * yq) % md;
}
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    for (j = 0; j < i; j++) {
      p[i][j] = a[i];
      p[j][i] = a[j];
      q[i][j] = q[j][i] = a[i] + a[j];
    }
  }
  rp = 0;
  rq = 1;
  for (i = 1; i < (1 << n); i++) {
    b[i] = b[i / 2] + (i & 1);
    cp[i] = zp = 1;
    cq[i] = zq = 1;
    for (m = j = 0; j < n; j++)
      if ((i >> j) & 1) {
        pos[m++] = j;
        for (k = 0; k < n; k++)
          if (((i >> k) & 1) == 0) mul(zp, zq, p[j][k], q[j][k]);
      }
    for (msk = (i & (i - 1)); msk > 0; msk = ((msk - 1) & i)) {
      curp = cp[msk];
      curq = cq[msk];
      for (j = 0; j < m; j++)
        if ((msk >> pos[j]) & 1)
          for (k = 0; k < m; k++)
            if (((msk >> pos[k]) & 1) == 0)
              mul(curp, curq, p[pos[j]][pos[k]], q[pos[j]][pos[k]]);
      sub(cp[i], cq[i], curp, curq);
    }
    inc(rp, rq, (((b[i] * cp[i]) % md) * zp) % md, (cq[i] * zq) % md);
  }
  printf("%lld\n", (rp * pw(rq, md - 2)) % md);
  return 0;
}

```