#include <bits/stdc++.h>
using namespace std;
void per() { cerr << endl; }
template <typename Head, typename... Tail>
void per(Head H, Tail... T) {
  cerr << H << ' ';
  per(T...);
}
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class U, class V>
istream& operator>>(istream& in, pair<U, V>& a) {
  return in >> a.first >> a.second;
}
template <typename W,
          typename T = typename enable_if<!is_same<W, string>::value,
                                          typename W::value_type>::type>
ostream& operator<<(ostream& out, const W& v) {
  out << "{ ";
  for (const auto& first : v) out << first << ", ";
  return out << '}';
}
template <class T>
void readArr(T from, T to) {
  for (auto i = from; i != to; ++i) cin >> *i;
}
mt19937 mrand(1337);
unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
unsigned long long myRand64() {
  return ((unsigned long long)myRand32() << 32) ^ myRand32();
}
const int mod = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void dec(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return a * (long long)b % mod; }
int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 14;
int dp[1 << N];
int pp[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto& first : v) cin >> first;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      pp[i][j] = mult(v[i], bp(v[i] + v[j], mod - 2));
    }
  }
  int ans = 0;
  for (int mask = 1; mask < (1 << n); ++mask) {
    int p = 1;
    vector<int> a, b;
    for (int i = 0; i < n; ++i) {
      if ((1 << i) & mask)
        a.push_back(i);
      else
        b.push_back(i);
    }
    for (auto first : a)
      for (auto second : b) p = mult(p, pp[first][second]);
    for (int s = mask; s; s = (s - 1) & mask) {
      if (s == mask) continue;
      int t = dp[mask ^ s];
      vector<int> c;
      for (int i = 0; i < n; ++i) {
        if ((1 << i) & s) c.push_back(i);
      }
      for (auto first : c)
        for (auto second : b) t = mult(t, pp[first][second]);
      dec(p, t);
    }
    dp[mask] = p;
    add(ans, mult(p, __builtin_popcount(mask)));
  }
  cout << 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;
void per() { cerr << endl; }
template <typename Head, typename... Tail>
void per(Head H, Tail... T) {
  cerr << H << ' ';
  per(T...);
}
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class U, class V>
istream& operator>>(istream& in, pair<U, V>& a) {
  return in >> a.first >> a.second;
}
template <typename W,
          typename T = typename enable_if<!is_same<W, string>::value,
                                          typename W::value_type>::type>
ostream& operator<<(ostream& out, const W& v) {
  out << "{ ";
  for (const auto& first : v) out << first << ", ";
  return out << '}';
}
template <class T>
void readArr(T from, T to) {
  for (auto i = from; i != to; ++i) cin >> *i;
}
mt19937 mrand(1337);
unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
unsigned long long myRand64() {
  return ((unsigned long long)myRand32() << 32) ^ myRand32();
}
const int mod = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void dec(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return a * (long long)b % mod; }
int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 14;
int dp[1 << N];
int pp[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto& first : v) cin >> first;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      pp[i][j] = mult(v[i], bp(v[i] + v[j], mod - 2));
    }
  }
  int ans = 0;
  for (int mask = 1; mask < (1 << n); ++mask) {
    int p = 1;
    vector<int> a, b;
    for (int i = 0; i < n; ++i) {
      if ((1 << i) & mask)
        a.push_back(i);
      else
        b.push_back(i);
    }
    for (auto first : a)
      for (auto second : b) p = mult(p, pp[first][second]);
    for (int s = mask; s; s = (s - 1) & mask) {
      if (s == mask) continue;
      int t = dp[mask ^ s];
      vector<int> c;
      for (int i = 0; i < n; ++i) {
        if ((1 << i) & s) c.push_back(i);
      }
      for (auto first : c)
        for (auto second : b) t = mult(t, pp[first][second]);
      dec(p, t);
    }
    dp[mask] = p;
    add(ans, mult(p, __builtin_popcount(mask)));
  }
  cout << ans;
}

```