#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 16;
const long long M = 5e6 + 5;
long long n, a[N], P[20005], DF[M], res = 0, Div[N][N], MaxM;
vector<pair<long long, long long> > lst;
void add(long long &a, long long b) { a = (a + b) % MOD; }
long long Pow(long long a, long long b) {
  if (b == 0) return 1;
  long long t = Pow(a, b / 2);
  if (b % 2 == 0) return t * t % MOD;
  return t * t % MOD * a % MOD;
}
long long check_bit(long long i, long long mask) {
  return (mask & (1 << i)) > 0;
}
long long Find(pair<long long, long long> t) {
  long long low = 0, high = lst.size() - 1;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (lst[mid].first < t.first)
      low = mid + 1;
    else {
      if (lst[mid].first > t.first)
        high = mid - 1;
      else {
        if (lst[mid].second > t.second)
          high = mid - 1;
        else {
          if (lst[mid].second < t.second)
            low = mid + 1;
          else
            return mid;
        }
      }
    }
  }
}
void Prepare() {
  for (long long i = 0; i <= n - 1; i++)
    for (long long j = i + 1; j <= n - 1; j++) {
      if (i == j) continue;
      Div[i][j] = Pow(a[i] + a[j], MOD - 2) % MOD;
      Div[j][i] = Div[i][j];
    }
  for (long long mask1 = 1; mask1 <= MaxM; mask1++) {
    for (long long mask2 = 0; mask2 <= (MaxM ^ mask1); mask2++) {
      if ((mask1 & mask2) != 0) continue;
      lst.push_back({mask1, mask2});
      long long num = __builtin_popcount(mask1),
                firstPos = __builtin_ctz(mask1);
      long long cur = lst.size() - 1;
      DF[cur] = 1;
      if (num == 1) {
        for (long long i = 0; i < n; i++) {
          if (check_bit(i, mask2)) {
            DF[cur] = DF[cur] * a[firstPos] % MOD * Div[firstPos][i] % MOD;
          }
        }
      } else {
        long long subMask = (1 << firstPos), ano_subMask = (mask1 ^ subMask);
        DF[cur] =
            DF[Find({subMask, mask2})] * DF[Find({ano_subMask, mask2})] % MOD;
      }
    }
  }
}
void Solve() {
  long long res = 0;
  for (long long mask = 1; mask <= MaxM; mask++) {
    long long num_Win = __builtin_popcount(mask);
    long long Lose = (MaxM ^ mask);
    P[mask] = 1;
    for (long long subMask = 1; subMask <= mask - 1; subMask++) {
      if ((mask & subMask) != subMask) continue;
      long long ano_subMask = (mask ^ subMask);
      P[mask] = (P[mask] - P[subMask] * DF[Find({subMask, ano_subMask})] % MOD +
                 MOD) %
                MOD;
    }
    add(res, P[mask] * DF[Find({mask, Lose})] % MOD * num_Win % MOD);
  }
  cout << res << '\n';
}
void in() {
  cin >> n;
  for (long long i = 0; i <= n - 1; i++) cin >> a[i];
  MaxM = (1 << n) - 1;
}
void process() {
  Prepare();
  Solve();
}
int main() {
  if (fopen("trash.inp", "r"))
    freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  in();
  process();
}

### Prompt

Please formulate 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;
const long long MOD = 1e9 + 7;
const long long N = 16;
const long long M = 5e6 + 5;
long long n, a[N], P[20005], DF[M], res = 0, Div[N][N], MaxM;
vector<pair<long long, long long> > lst;
void add(long long &a, long long b) { a = (a + b) % MOD; }
long long Pow(long long a, long long b) {
  if (b == 0) return 1;
  long long t = Pow(a, b / 2);
  if (b % 2 == 0) return t * t % MOD;
  return t * t % MOD * a % MOD;
}
long long check_bit(long long i, long long mask) {
  return (mask & (1 << i)) > 0;
}
long long Find(pair<long long, long long> t) {
  long long low = 0, high = lst.size() - 1;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (lst[mid].first < t.first)
      low = mid + 1;
    else {
      if (lst[mid].first > t.first)
        high = mid - 1;
      else {
        if (lst[mid].second > t.second)
          high = mid - 1;
        else {
          if (lst[mid].second < t.second)
            low = mid + 1;
          else
            return mid;
        }
      }
    }
  }
}
void Prepare() {
  for (long long i = 0; i <= n - 1; i++)
    for (long long j = i + 1; j <= n - 1; j++) {
      if (i == j) continue;
      Div[i][j] = Pow(a[i] + a[j], MOD - 2) % MOD;
      Div[j][i] = Div[i][j];
    }
  for (long long mask1 = 1; mask1 <= MaxM; mask1++) {
    for (long long mask2 = 0; mask2 <= (MaxM ^ mask1); mask2++) {
      if ((mask1 & mask2) != 0) continue;
      lst.push_back({mask1, mask2});
      long long num = __builtin_popcount(mask1),
                firstPos = __builtin_ctz(mask1);
      long long cur = lst.size() - 1;
      DF[cur] = 1;
      if (num == 1) {
        for (long long i = 0; i < n; i++) {
          if (check_bit(i, mask2)) {
            DF[cur] = DF[cur] * a[firstPos] % MOD * Div[firstPos][i] % MOD;
          }
        }
      } else {
        long long subMask = (1 << firstPos), ano_subMask = (mask1 ^ subMask);
        DF[cur] =
            DF[Find({subMask, mask2})] * DF[Find({ano_subMask, mask2})] % MOD;
      }
    }
  }
}
void Solve() {
  long long res = 0;
  for (long long mask = 1; mask <= MaxM; mask++) {
    long long num_Win = __builtin_popcount(mask);
    long long Lose = (MaxM ^ mask);
    P[mask] = 1;
    for (long long subMask = 1; subMask <= mask - 1; subMask++) {
      if ((mask & subMask) != subMask) continue;
      long long ano_subMask = (mask ^ subMask);
      P[mask] = (P[mask] - P[subMask] * DF[Find({subMask, ano_subMask})] % MOD +
                 MOD) %
                MOD;
    }
    add(res, P[mask] * DF[Find({mask, Lose})] % MOD * num_Win % MOD);
  }
  cout << res << '\n';
}
void in() {
  cin >> n;
  for (long long i = 0; i <= n - 1; i++) cin >> a[i];
  MaxM = (1 << n) - 1;
}
void process() {
  Prepare();
  Solve();
}
int main() {
  if (fopen("trash.inp", "r"))
    freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  in();
  process();
}

```