#include <bits/stdc++.h>
using namespace std;
long long mmod(long long x, long long y) {
  if (x >= 0LL)
    return x % y;
  else
    return (y - (-x) % y) % y;
}
long long ddiv(long long x, long long y) {
  x -= mmod(x, y);
  return x / y;
}
long long binpow(long long a, long long b, long long mod) {
  if (a == 0) {
    return 0;
  }
  long long pows[65];
  pows[0] = a;
  for (int i = 1; i < 65; ++i) {
    pows[i] = mmod(pows[i - 1] * pows[i - 1], mod);
  }
  long long ans = 1;
  for (int i = 0; b != 0; b >>= 1, ++i) {
    if (b % 2 == 1) {
      ans = mmod(ans * pows[i], mod);
    }
  }
  return ans;
}
long long GCD(long long a, long long b) {
  while (a != 0) {
    b = b % a;
    swap(a, b);
  }
  return b;
}
const int M = 1000000007;
int bmRange;
vector<int> a;
struct frac {
  long long num;
  long long denom;
  frac() : num(0), denom(1) {}
  frac(long long n, long long d) : num(n), denom(d) {}
  frac operator*(const frac& other) const {
    return {mmod(num * other.num, M), mmod(denom * other.denom, M)};
  }
  frac operator+(const frac& other) const {
    return {mmod(num * other.denom + denom * other.num, M),
            mmod(denom * other.denom, M)};
  }
  frac operator-() const { return {mmod(denom - num, M), denom}; }
  long long get() const { return mmod(num * binpow(denom, M - 2, M), M); }
};
vector<frac> probs;
frac get_prob(int submask, int bitmask) {
  frac res(1, 1);
  for (int i = 0; i < 14; ++i)
    for (int j = 0; j < 14; ++j) {
      if ((submask >> i) & 1 && (bitmask >> j) & 1 && !((submask >> j) & 1)) {
        res = res * frac(a[i], a[i] + a[j]);
      }
    }
  return res;
}
frac res{0, 1};
void apply(int bitmask) {
  for (int submask = bitmask & (bitmask - 1); submask != 0;
       submask = bitmask & (submask - 1)) {
    probs[bitmask] =
        probs[bitmask] + (-probs[submask]) * get_prob(submask, bitmask);
  }
  int memc = 0;
  for (int i = 0; i < 14; ++i) {
    if ((bitmask >> i) & 1) ++memc;
  }
  res =
      res + (-probs[bitmask]) * get_prob(bitmask, bmRange - 1) * frac(memc, 1);
}
void solve() {
  int n;
  cin >> n;
  bmRange = (1 << n);
  a.resize(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  probs.resize(bmRange, {0, 1});
  for (int i = 0; i < bmRange; ++i) apply(i);
  cout << res.get() << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  for (int k = 0; k < t; ++k) 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;
long long mmod(long long x, long long y) {
  if (x >= 0LL)
    return x % y;
  else
    return (y - (-x) % y) % y;
}
long long ddiv(long long x, long long y) {
  x -= mmod(x, y);
  return x / y;
}
long long binpow(long long a, long long b, long long mod) {
  if (a == 0) {
    return 0;
  }
  long long pows[65];
  pows[0] = a;
  for (int i = 1; i < 65; ++i) {
    pows[i] = mmod(pows[i - 1] * pows[i - 1], mod);
  }
  long long ans = 1;
  for (int i = 0; b != 0; b >>= 1, ++i) {
    if (b % 2 == 1) {
      ans = mmod(ans * pows[i], mod);
    }
  }
  return ans;
}
long long GCD(long long a, long long b) {
  while (a != 0) {
    b = b % a;
    swap(a, b);
  }
  return b;
}
const int M = 1000000007;
int bmRange;
vector<int> a;
struct frac {
  long long num;
  long long denom;
  frac() : num(0), denom(1) {}
  frac(long long n, long long d) : num(n), denom(d) {}
  frac operator*(const frac& other) const {
    return {mmod(num * other.num, M), mmod(denom * other.denom, M)};
  }
  frac operator+(const frac& other) const {
    return {mmod(num * other.denom + denom * other.num, M),
            mmod(denom * other.denom, M)};
  }
  frac operator-() const { return {mmod(denom - num, M), denom}; }
  long long get() const { return mmod(num * binpow(denom, M - 2, M), M); }
};
vector<frac> probs;
frac get_prob(int submask, int bitmask) {
  frac res(1, 1);
  for (int i = 0; i < 14; ++i)
    for (int j = 0; j < 14; ++j) {
      if ((submask >> i) & 1 && (bitmask >> j) & 1 && !((submask >> j) & 1)) {
        res = res * frac(a[i], a[i] + a[j]);
      }
    }
  return res;
}
frac res{0, 1};
void apply(int bitmask) {
  for (int submask = bitmask & (bitmask - 1); submask != 0;
       submask = bitmask & (submask - 1)) {
    probs[bitmask] =
        probs[bitmask] + (-probs[submask]) * get_prob(submask, bitmask);
  }
  int memc = 0;
  for (int i = 0; i < 14; ++i) {
    if ((bitmask >> i) & 1) ++memc;
  }
  res =
      res + (-probs[bitmask]) * get_prob(bitmask, bmRange - 1) * frac(memc, 1);
}
void solve() {
  int n;
  cin >> n;
  bmRange = (1 << n);
  a.resize(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  probs.resize(bmRange, {0, 1});
  for (int i = 0; i < bmRange; ++i) apply(i);
  cout << res.get() << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  for (int k = 0; k < t; ++k) solve();
  return 0;
}

```