#include <bits/stdc++.h>
using namespace std;
const int INFi = 1e9 + 5;
const int md = 1e9 + 7;
const long long INF = 2e18;
const int maxN = 5e5 + 1;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; }
inline int add(const int &a, const int &b) {
  return a + b >= md ? a + b - md : a + b;
}
inline int sub(const int &a, const int &b) {
  return a - b < 0 ? a - b + md : a - b;
}
inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
int binpow(int a, int b) {
  if (b <= 0) return 1;
  if (b % 2) return mul(a, binpow(a, b - 1));
  int m = binpow(a, b / 2);
  return mul(m, m);
}
int rev(int a) { return binpow(a, md - 2); }
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  vector<vector<int>> prl(n, vector<int>(1 << n));
  for (int i = 0; i < (n); ++i) {
    for (int mask = 0; mask < (1 << n); ++mask) {
      if (mask & (1 << i)) {
        prl[i][mask] = 0;
        continue;
      }
      prl[i][mask] = 1;
      for (int j = 0; j < (n); ++j) {
        if ((1 << j) & mask) {
          prl[i][mask] = mul(prl[i][mask], mul(a[i], rev(add(a[i], a[j]))));
        }
      }
    }
  }
  vector<int> pw3(n + 1);
  pw3[0] = 1;
  for (int i = 1; i <= n; ++i) pw3[i] = pw3[i - 1] * 3;
  vector<int> dp(pw3[n]);
  vector<int> prwin(pw3[n]), prlose(pw3[n]);
  vector<int> to(1 << n);
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int i = 0; i < (n); ++i) {
      if (mask & (1 << i)) to[mask] += pw3[i];
    }
  }
  for (int mask3 = 0; mask3 < (pw3[n]); ++mask3) {
    int mask1 = 0, mask2 = 0;
    {
      int c = mask3;
      for (int i = 0; i < (n); ++i) {
        int x = c % 3;
        c /= 3;
        if (x == 1) mask1 |= (1 << i);
        if (x == 2) mask2 |= (1 << i);
      }
    }
    prwin[mask3] = 1;
    prlose[mask3] = 1;
    for (int i = 0; i < (n); ++i) {
      if ((1 << i) & mask2) {
        prlose[mask3] = mul(prlose[mask3], prl[i][mask1]);
        prwin[mask3] = mul(prwin[mask3], sub(1, prl[i][mask1]));
      }
    }
  }
  for (int i = 0; i < (n); ++i) dp[pw3[i]] = 1;
  int ans = 0;
  for (int mask3 = 0; mask3 < (pw3[n]); ++mask3) {
    int mask1 = 0, mask2 = 0, mask0 = 0;
    {
      int c = mask3;
      for (int i = 0; i < (n); ++i) {
        int x = c % 3;
        c /= 3;
        if (x) mask1 |= (1 << i);
        if (x == 1) mask2 |= (1 << i);
        if (x == 0) mask0 |= (1 << i);
      }
    }
    for (int submask = mask0; submask > 0; submask = (submask - 1) & mask0) {
      int cur = dp[mask3];
      int nxt = to[mask1] * 2 + to[submask];
      cur = mul(cur, prwin[to[submask] * 2 + to[mask2]]);
      cur = mul(cur, prlose[to[submask] * 2 + to[mask1 ^ mask2]]);
      dp[nxt] = add(dp[nxt], cur);
    }
    if (!mask0) ans = add(ans, dp[mask3]);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tests = 1;
  for (int _ = 0; _ < (tests); ++_) {
    solve();
  }
  return 0;
}

### 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 int INFi = 1e9 + 5;
const int md = 1e9 + 7;
const long long INF = 2e18;
const int maxN = 5e5 + 1;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; }
inline int add(const int &a, const int &b) {
  return a + b >= md ? a + b - md : a + b;
}
inline int sub(const int &a, const int &b) {
  return a - b < 0 ? a - b + md : a - b;
}
inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
int binpow(int a, int b) {
  if (b <= 0) return 1;
  if (b % 2) return mul(a, binpow(a, b - 1));
  int m = binpow(a, b / 2);
  return mul(m, m);
}
int rev(int a) { return binpow(a, md - 2); }
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  vector<vector<int>> prl(n, vector<int>(1 << n));
  for (int i = 0; i < (n); ++i) {
    for (int mask = 0; mask < (1 << n); ++mask) {
      if (mask & (1 << i)) {
        prl[i][mask] = 0;
        continue;
      }
      prl[i][mask] = 1;
      for (int j = 0; j < (n); ++j) {
        if ((1 << j) & mask) {
          prl[i][mask] = mul(prl[i][mask], mul(a[i], rev(add(a[i], a[j]))));
        }
      }
    }
  }
  vector<int> pw3(n + 1);
  pw3[0] = 1;
  for (int i = 1; i <= n; ++i) pw3[i] = pw3[i - 1] * 3;
  vector<int> dp(pw3[n]);
  vector<int> prwin(pw3[n]), prlose(pw3[n]);
  vector<int> to(1 << n);
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int i = 0; i < (n); ++i) {
      if (mask & (1 << i)) to[mask] += pw3[i];
    }
  }
  for (int mask3 = 0; mask3 < (pw3[n]); ++mask3) {
    int mask1 = 0, mask2 = 0;
    {
      int c = mask3;
      for (int i = 0; i < (n); ++i) {
        int x = c % 3;
        c /= 3;
        if (x == 1) mask1 |= (1 << i);
        if (x == 2) mask2 |= (1 << i);
      }
    }
    prwin[mask3] = 1;
    prlose[mask3] = 1;
    for (int i = 0; i < (n); ++i) {
      if ((1 << i) & mask2) {
        prlose[mask3] = mul(prlose[mask3], prl[i][mask1]);
        prwin[mask3] = mul(prwin[mask3], sub(1, prl[i][mask1]));
      }
    }
  }
  for (int i = 0; i < (n); ++i) dp[pw3[i]] = 1;
  int ans = 0;
  for (int mask3 = 0; mask3 < (pw3[n]); ++mask3) {
    int mask1 = 0, mask2 = 0, mask0 = 0;
    {
      int c = mask3;
      for (int i = 0; i < (n); ++i) {
        int x = c % 3;
        c /= 3;
        if (x) mask1 |= (1 << i);
        if (x == 1) mask2 |= (1 << i);
        if (x == 0) mask0 |= (1 << i);
      }
    }
    for (int submask = mask0; submask > 0; submask = (submask - 1) & mask0) {
      int cur = dp[mask3];
      int nxt = to[mask1] * 2 + to[submask];
      cur = mul(cur, prwin[to[submask] * 2 + to[mask2]]);
      cur = mul(cur, prlose[to[submask] * 2 + to[mask1 ^ mask2]]);
      dp[nxt] = add(dp[nxt], cur);
    }
    if (!mask0) ans = add(ans, dp[mask3]);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tests = 1;
  for (int _ = 0; _ < (tests); ++_) {
    solve();
  }
  return 0;
}

```