#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <int MOD>
struct Modular {
  int val;
  Modular() : val(0){};
  Modular(int _val) : val(_val % MOD){};
  void norm() {
    val %= MOD;
    if (val < 0) val += MOD;
  }
  Modular& operator+=(const Modular& other) {
    val += other.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  Modular operator+(const Modular& other) const {
    auto tmp(*this);
    tmp += other;
    return tmp;
  }
  Modular& operator-=(const Modular& other) {
    val -= other.val;
    if (val < 0) val += MOD;
    return *this;
  }
  Modular operator-(const Modular& other) const {
    auto tmp(*this);
    tmp -= other;
    return tmp;
  }
  Modular& operator*=(const Modular& other) {
    val = (val * 1LL * other.val) % MOD;
    return *this;
  }
  Modular operator*(const Modular& other) const {
    auto tmp(*this);
    tmp *= other;
    return tmp;
  }
  Modular bin_pow(int n) const {
    Modular res(1);
    Modular a = *this;
    for (; n; n >>= 1, a *= a)
      if (n & 1) res *= a;
    return res;
  }
  Modular inv() const { return this->bin_pow(MOD - 2); }
  Modular& operator/=(const Modular& other) { return *this *= other.inv(); }
  Modular operator/(const Modular& other) const { return *this * other.inv(); }
};
template <int MOD>
std::ostream& operator<<(std::ostream& out, const Modular<MOD>& x) {
  out << x.val;
  return out;
}
const int MOD = (int)1e9 + 7;
using Mint = Modular<MOD>;
int bit(int mask, int k) { return (mask >> k) & 1; }
const int N = 14;
int a[N];
Mint p[N][1 << N];
Mint scc[1 << N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  auto getp = [&](int i, int j) { return Mint(a[i]) / Mint(a[i] + a[j]); };
  auto getpp = [&](int mask1, int mask2) {
    Mint res(1);
    for (int i = 0; i < n; ++i) {
      if (bit(mask1, i)) {
        res *= p[i][mask2];
      }
    }
    return res;
  };
  for (int i = 0; i < n; ++i) {
    for (int mask = 0; mask < (1 << n); ++mask) {
      if (bit(mask, i)) continue;
      p[i][mask] = Mint(1);
      for (int j = 0; j < n; ++j) {
        if (bit(mask, j)) {
          p[i][mask] *= getp(i, j);
        }
      }
    }
  }
  Mint ans(0);
  for (int mask = 0; mask < (1 << n); ++mask) {
    scc[mask] = Mint(1);
    if (__builtin_popcount(mask) > 1) {
      for (int smask = (mask - 1) & mask; smask; smask = (smask - 1) & mask) {
        scc[mask] -= scc[smask] * getpp(smask, mask ^ smask);
      }
    }
    ans += Mint(__builtin_popcount(mask)) * scc[mask] *
           getpp(mask, mask ^ ((1 << n) - 1));
  }
  cout << ans << '\n';
}

### Prompt

Your task is to create 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <int MOD>
struct Modular {
  int val;
  Modular() : val(0){};
  Modular(int _val) : val(_val % MOD){};
  void norm() {
    val %= MOD;
    if (val < 0) val += MOD;
  }
  Modular& operator+=(const Modular& other) {
    val += other.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  Modular operator+(const Modular& other) const {
    auto tmp(*this);
    tmp += other;
    return tmp;
  }
  Modular& operator-=(const Modular& other) {
    val -= other.val;
    if (val < 0) val += MOD;
    return *this;
  }
  Modular operator-(const Modular& other) const {
    auto tmp(*this);
    tmp -= other;
    return tmp;
  }
  Modular& operator*=(const Modular& other) {
    val = (val * 1LL * other.val) % MOD;
    return *this;
  }
  Modular operator*(const Modular& other) const {
    auto tmp(*this);
    tmp *= other;
    return tmp;
  }
  Modular bin_pow(int n) const {
    Modular res(1);
    Modular a = *this;
    for (; n; n >>= 1, a *= a)
      if (n & 1) res *= a;
    return res;
  }
  Modular inv() const { return this->bin_pow(MOD - 2); }
  Modular& operator/=(const Modular& other) { return *this *= other.inv(); }
  Modular operator/(const Modular& other) const { return *this * other.inv(); }
};
template <int MOD>
std::ostream& operator<<(std::ostream& out, const Modular<MOD>& x) {
  out << x.val;
  return out;
}
const int MOD = (int)1e9 + 7;
using Mint = Modular<MOD>;
int bit(int mask, int k) { return (mask >> k) & 1; }
const int N = 14;
int a[N];
Mint p[N][1 << N];
Mint scc[1 << N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  auto getp = [&](int i, int j) { return Mint(a[i]) / Mint(a[i] + a[j]); };
  auto getpp = [&](int mask1, int mask2) {
    Mint res(1);
    for (int i = 0; i < n; ++i) {
      if (bit(mask1, i)) {
        res *= p[i][mask2];
      }
    }
    return res;
  };
  for (int i = 0; i < n; ++i) {
    for (int mask = 0; mask < (1 << n); ++mask) {
      if (bit(mask, i)) continue;
      p[i][mask] = Mint(1);
      for (int j = 0; j < n; ++j) {
        if (bit(mask, j)) {
          p[i][mask] *= getp(i, j);
        }
      }
    }
  }
  Mint ans(0);
  for (int mask = 0; mask < (1 << n); ++mask) {
    scc[mask] = Mint(1);
    if (__builtin_popcount(mask) > 1) {
      for (int smask = (mask - 1) & mask; smask; smask = (smask - 1) & mask) {
        scc[mask] -= scc[smask] * getpp(smask, mask ^ smask);
      }
    }
    ans += Mint(__builtin_popcount(mask)) * scc[mask] *
           getpp(mask, mask ^ ((1 << n) - 1));
  }
  cout << ans << '\n';
}

```