#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  timeEnd();
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
class InputReader {
 public:
  InputReader() {
    input_file = stdin;
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXN = 14;
const int MOD = 1000000007;
int raiseToPower(int base, int power) {
  int answer = 1;
  while (power) {
    if (power % 2) answer = (1LL * answer * base) % MOD;
    base = (1LL * base * base) % MOD;
    power /= 2;
  }
  return answer;
}
int v[MAXN];
int mul(int a, int b) { return 1LL * a * b % MOD; }
int inv(int x) { return raiseToPower(x, MOD - 2); }
int win(int i, int j) { return mul(v[i], inv(v[i] + v[j])); }
int lose(int i, int j) { return mul(v[j], inv(v[i] + v[j])); }
void add(int &x, int y) {
  x += y;
  if (x >= MOD) {
    x -= MOD;
  }
}
void sub(int &x, int y) {
  x -= y;
  if (x < 0) {
    x += MOD;
  }
}
int beat[MAXN][1 << MAXN];
void precompute(int n) {
  for (int x = 0; x < n; x++) {
    for (int mask = 0; mask < (1 << n); mask++) {
      beat[x][mask] = 1;
      for (int i = 0; i < n; i++) {
        if (((mask >> i) & 1) && i != x) {
          beat[x][mask] = mul(beat[x][mask], win(x, i));
        }
      }
    }
  }
}
bool seen[1 << MAXN];
int pr[1 << MAXN], dp[1 << MAXN];
int sep(int s1, int s2, int n) {
  int p = 1;
  for (int i = 0; i < n; i++) {
    if ((s1 >> i) & 1) {
      p = mul(p, beat[i][s2]);
    }
  }
  return p;
}
void compute(int mask, int n) {
  if (seen[mask]) {
    return;
  }
  seen[mask] = true;
  if (__builtin_popcount(mask) == 1) {
    pr[mask] = dp[mask] = 1;
    return;
  }
  pr[mask] = 1;
  for (int s1 = (mask - 1) & mask; s1 != 0; s1 = (s1 - 1) & mask) {
    int s2 = mask ^ s1;
    assert((s1 & mask) == s1);
    assert((s2 & mask) == s2);
    assert((s1 & s2) == 0);
    assert((s1 | s2) == mask);
    compute(s1, n);
    compute(s2, n);
    int p = mul(sep(s1, s2, n), pr[s1]);
    sub(pr[mask], p);
    add(dp[mask], mul(p, __builtin_popcount(s1)));
  }
  add(dp[mask], mul(pr[mask], __builtin_popcount(mask)));
}
int main() {
  timeBegin();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  srand(time(0));
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  precompute(n);
  compute((1 << n) - 1, n);
  cout << dp[(1 << n) - 1] << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp 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 target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  timeEnd();
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
class InputReader {
 public:
  InputReader() {
    input_file = stdin;
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  InputReader(const char *file_name) {
    input_file = fopen(file_name, "r");
    cursor = 0;
    fread(buffer, SIZE, 1, input_file);
  }
  inline InputReader &operator>>(int &n) {
    while ((buffer[cursor] < '0' || buffer[cursor] > '9') &&
           buffer[cursor] != '-') {
      advance();
    }
    int sign = 1;
    if (buffer[cursor] == '-') {
      sign = -1;
      advance();
    }
    n = 0;
    while ('0' <= buffer[cursor] && buffer[cursor] <= '9') {
      n = n * 10 + buffer[cursor] - '0';
      advance();
    }
    n *= sign;
    return *this;
  }

 private:
  FILE *input_file;
  static const int SIZE = 1 << 17;
  int cursor;
  char buffer[SIZE];
  inline void advance() {
    ++cursor;
    if (cursor == SIZE) {
      cursor = 0;
      fread(buffer, SIZE, 1, input_file);
    }
  }
};
const int MAXN = 14;
const int MOD = 1000000007;
int raiseToPower(int base, int power) {
  int answer = 1;
  while (power) {
    if (power % 2) answer = (1LL * answer * base) % MOD;
    base = (1LL * base * base) % MOD;
    power /= 2;
  }
  return answer;
}
int v[MAXN];
int mul(int a, int b) { return 1LL * a * b % MOD; }
int inv(int x) { return raiseToPower(x, MOD - 2); }
int win(int i, int j) { return mul(v[i], inv(v[i] + v[j])); }
int lose(int i, int j) { return mul(v[j], inv(v[i] + v[j])); }
void add(int &x, int y) {
  x += y;
  if (x >= MOD) {
    x -= MOD;
  }
}
void sub(int &x, int y) {
  x -= y;
  if (x < 0) {
    x += MOD;
  }
}
int beat[MAXN][1 << MAXN];
void precompute(int n) {
  for (int x = 0; x < n; x++) {
    for (int mask = 0; mask < (1 << n); mask++) {
      beat[x][mask] = 1;
      for (int i = 0; i < n; i++) {
        if (((mask >> i) & 1) && i != x) {
          beat[x][mask] = mul(beat[x][mask], win(x, i));
        }
      }
    }
  }
}
bool seen[1 << MAXN];
int pr[1 << MAXN], dp[1 << MAXN];
int sep(int s1, int s2, int n) {
  int p = 1;
  for (int i = 0; i < n; i++) {
    if ((s1 >> i) & 1) {
      p = mul(p, beat[i][s2]);
    }
  }
  return p;
}
void compute(int mask, int n) {
  if (seen[mask]) {
    return;
  }
  seen[mask] = true;
  if (__builtin_popcount(mask) == 1) {
    pr[mask] = dp[mask] = 1;
    return;
  }
  pr[mask] = 1;
  for (int s1 = (mask - 1) & mask; s1 != 0; s1 = (s1 - 1) & mask) {
    int s2 = mask ^ s1;
    assert((s1 & mask) == s1);
    assert((s2 & mask) == s2);
    assert((s1 & s2) == 0);
    assert((s1 | s2) == mask);
    compute(s1, n);
    compute(s2, n);
    int p = mul(sep(s1, s2, n), pr[s1]);
    sub(pr[mask], p);
    add(dp[mask], mul(p, __builtin_popcount(s1)));
  }
  add(dp[mask], mul(pr[mask], __builtin_popcount(mask)));
}
int main() {
  timeBegin();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  srand(time(0));
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  precompute(n);
  compute((1 << n) - 1, n);
  cout << dp[(1 << n) - 1] << "\n";
  return 0;
}

```