#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
struct mi {
  typedef decay<decltype(MOD)>::type T;
  T v;
  explicit operator T() const { return v; }
  mi() { v = 0; }
  mi(long long _v) {
    v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
    if (v < 0) v += MOD;
  }
  friend bool operator==(const mi& a, const mi& b) { return a.v == b.v; }
  friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
  friend bool operator<(const mi& a, const mi& b) { return a.v < b.v; }
  mi& operator+=(const mi& m) {
    if ((v += m.v) >= MOD) v -= MOD;
    return *this;
  }
  mi& operator-=(const mi& m) {
    if ((v -= m.v) < 0) v += MOD;
    return *this;
  }
  mi& operator*=(const mi& m) {
    v = (long long)v * m.v % MOD;
    return *this;
  }
  mi& operator/=(const mi& m) { return (*this) *= inv(m); }
  friend mi pow(mi a, long long p) {
    mi ans = 1;
    assert(p >= 0);
    for (; p; p /= 2, a *= a)
      if (p & 1) ans *= a;
    return ans;
  }
  friend mi inv(const mi& a) {
    assert(a.v != 0);
    return pow(a, MOD - 2);
  }
  mi operator-() const { return mi(-v); }
  mi& operator++() { return *this += 1; }
  mi& operator--() { return *this -= 1; }
  friend mi operator+(mi a, const mi& b) { return a += b; }
  friend mi operator-(mi a, const mi& b) { return a -= b; }
  friend mi operator*(mi a, const mi& b) { return a *= b; }
  friend mi operator/(mi a, const mi& b) { return a /= b; }
};
const int N = 15, M = (1 << 14) + 5;
map<vector<int>, int> mp;
void pre(vector<int> vs) {}
int n, a[N];
int bcnt[M];
vector<int> bt[M];
mi win[N][N];
mi dp[M];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> a[i];
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      win[i][j] = mi(a[i]) / (mi(a[i]) + mi(a[j]));
    }
  }
  for (int i = 1; i < (1 << n); ++i) bcnt[i] = bcnt[i / 2] + (i & 1);
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int b = 0; b < (n); ++b)
      if (mask >> b & 1) bt[mask].push_back(b);
  }
  mi fres = 0;
  for (int mask = 1; mask < (1 << n); ++mask) {
    dp[mask] = 1;
    for (int sub = mask;; sub = (sub - 1) & mask) {
      if (sub == mask) continue;
      int rem = mask ^ sub;
      mi s = dp[sub];
      for (int b : bt[sub]) {
        for (int r : bt[rem]) {
          s *= win[b][r];
        }
      }
      dp[mask] -= s;
      if (sub == 0) break;
    }
    mi s = dp[mask];
    int rem = ((1 << n) - 1) ^ mask;
    for (int b : bt[mask]) {
      for (int r : bt[rem]) {
        s *= win[b][r];
      }
    }
    fres += s * bcnt[mask];
  }
  cout << int(fres) << '\n';
}

### Prompt

Create a solution in Cpp for 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;
const int MOD = (int)1e9 + 7;
struct mi {
  typedef decay<decltype(MOD)>::type T;
  T v;
  explicit operator T() const { return v; }
  mi() { v = 0; }
  mi(long long _v) {
    v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
    if (v < 0) v += MOD;
  }
  friend bool operator==(const mi& a, const mi& b) { return a.v == b.v; }
  friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
  friend bool operator<(const mi& a, const mi& b) { return a.v < b.v; }
  mi& operator+=(const mi& m) {
    if ((v += m.v) >= MOD) v -= MOD;
    return *this;
  }
  mi& operator-=(const mi& m) {
    if ((v -= m.v) < 0) v += MOD;
    return *this;
  }
  mi& operator*=(const mi& m) {
    v = (long long)v * m.v % MOD;
    return *this;
  }
  mi& operator/=(const mi& m) { return (*this) *= inv(m); }
  friend mi pow(mi a, long long p) {
    mi ans = 1;
    assert(p >= 0);
    for (; p; p /= 2, a *= a)
      if (p & 1) ans *= a;
    return ans;
  }
  friend mi inv(const mi& a) {
    assert(a.v != 0);
    return pow(a, MOD - 2);
  }
  mi operator-() const { return mi(-v); }
  mi& operator++() { return *this += 1; }
  mi& operator--() { return *this -= 1; }
  friend mi operator+(mi a, const mi& b) { return a += b; }
  friend mi operator-(mi a, const mi& b) { return a -= b; }
  friend mi operator*(mi a, const mi& b) { return a *= b; }
  friend mi operator/(mi a, const mi& b) { return a /= b; }
};
const int N = 15, M = (1 << 14) + 5;
map<vector<int>, int> mp;
void pre(vector<int> vs) {}
int n, a[N];
int bcnt[M];
vector<int> bt[M];
mi win[N][N];
mi dp[M];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> a[i];
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      win[i][j] = mi(a[i]) / (mi(a[i]) + mi(a[j]));
    }
  }
  for (int i = 1; i < (1 << n); ++i) bcnt[i] = bcnt[i / 2] + (i & 1);
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int b = 0; b < (n); ++b)
      if (mask >> b & 1) bt[mask].push_back(b);
  }
  mi fres = 0;
  for (int mask = 1; mask < (1 << n); ++mask) {
    dp[mask] = 1;
    for (int sub = mask;; sub = (sub - 1) & mask) {
      if (sub == mask) continue;
      int rem = mask ^ sub;
      mi s = dp[sub];
      for (int b : bt[sub]) {
        for (int r : bt[rem]) {
          s *= win[b][r];
        }
      }
      dp[mask] -= s;
      if (sub == 0) break;
    }
    mi s = dp[mask];
    int rem = ((1 << n) - 1) ^ mask;
    for (int b : bt[mask]) {
      for (int r : bt[rem]) {
        s *= win[b][r];
      }
    }
    fres += s * bcnt[mask];
  }
  cout << int(fres) << '\n';
}

```