#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
const long long N = 15;
const long long M = 2e6 + 5;
const long long SZ = 550;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const long long inf = 0x3f3f3f3f3f3f3f;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void wt(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) wt(x / 10), putchar(10);
  putchar(x % 10 + '0');
}
long long ksm(long long x, long long y = mod - 2, long long z = mod) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
long long inv[M], fac[M], ifc[M];
void Init(long long n) {
  inv[1] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifc[0] = 1;
  for (long long i = 1; i <= n; i++) ifc[i] = 1ll * ifc[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m) {
  if (m < 0 || m > n) return 0;
  return 1ll * fac[n] * ifc[m] % mod * ifc[n - m] % mod;
}
long long n, ans;
long long a[N], g[N][N];
long long f[1 << 14];
vector<long long> v1, v2;
signed main() {
  Init(M - 1);
  n = read();
  for (long long i = 0; i < n; i++) a[i] = read();
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) g[i][j] = a[i] * inv[a[i] + a[j]] % mod;
  for (long long S = 1; S < (1 << n); S++) {
    f[S] = 1;
    for (long long T = (S - 1) & S; T; T = (T - 1) & S) {
      long long X = S - T;
      v1.clear();
      v2.clear();
      for (long long i = 0; i < n; i++) {
        if (T >> i & 1) v1.push_back(i);
        if (X >> i & 1) v2.push_back(i);
      }
      long long P = 1;
      for (long long i : v1)
        for (long long j : v2) P = P * g[i][j] % mod;
      f[S] = (f[S] - P * f[T] % mod + mod) % mod;
    }
  }
  for (long long S = 1; S < (1 << n); S++) {
    v1.clear();
    v2.clear();
    for (long long i = 0; i < n; i++)
      if (S >> i & 1)
        v1.push_back(i);
      else
        v2.push_back(i);
    long long P = 1;
    for (long long i : v1)
      for (long long j : v2) P = P * g[i][j] % mod;
    ans = (ans + P * f[S] % mod * v1.size()) % mod;
  }
  cout << (ans + mod) % mod << "\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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
const long long N = 15;
const long long M = 2e6 + 5;
const long long SZ = 550;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const long long inf = 0x3f3f3f3f3f3f3f;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void wt(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) wt(x / 10), putchar(10);
  putchar(x % 10 + '0');
}
long long ksm(long long x, long long y = mod - 2, long long z = mod) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
long long inv[M], fac[M], ifc[M];
void Init(long long n) {
  inv[1] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifc[0] = 1;
  for (long long i = 1; i <= n; i++) ifc[i] = 1ll * ifc[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m) {
  if (m < 0 || m > n) return 0;
  return 1ll * fac[n] * ifc[m] % mod * ifc[n - m] % mod;
}
long long n, ans;
long long a[N], g[N][N];
long long f[1 << 14];
vector<long long> v1, v2;
signed main() {
  Init(M - 1);
  n = read();
  for (long long i = 0; i < n; i++) a[i] = read();
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) g[i][j] = a[i] * inv[a[i] + a[j]] % mod;
  for (long long S = 1; S < (1 << n); S++) {
    f[S] = 1;
    for (long long T = (S - 1) & S; T; T = (T - 1) & S) {
      long long X = S - T;
      v1.clear();
      v2.clear();
      for (long long i = 0; i < n; i++) {
        if (T >> i & 1) v1.push_back(i);
        if (X >> i & 1) v2.push_back(i);
      }
      long long P = 1;
      for (long long i : v1)
        for (long long j : v2) P = P * g[i][j] % mod;
      f[S] = (f[S] - P * f[T] % mod + mod) % mod;
    }
  }
  for (long long S = 1; S < (1 << n); S++) {
    v1.clear();
    v2.clear();
    for (long long i = 0; i < n; i++)
      if (S >> i & 1)
        v1.push_back(i);
      else
        v2.push_back(i);
    long long P = 1;
    for (long long i : v1)
      for (long long j : v2) P = P * g[i][j] % mod;
    ans = (ans + P * f[S] % mod * v1.size()) % mod;
  }
  cout << (ans + mod) % mod << "\n";
}

```