#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
template <typename T>
bool chmax(T &x, const T &y) {
  return (x < y) ? (x = y, true) : false;
};
template <typename T>
bool chmin(T &x, const T &y) {
  return (x > y) ? (x = y, true) : false;
};
struct io_setup {
  io_setup() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(15);
  }
} io_setup;
template <int mod>
struct Mod_Int {
  int x;
  Mod_Int() : x(0) {}
  Mod_Int(long long y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  static int get_mod() { return mod; }
  Mod_Int &operator+=(const Mod_Int &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  Mod_Int &operator-=(const Mod_Int &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  Mod_Int &operator*=(const Mod_Int &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  Mod_Int &operator/=(const Mod_Int &p) {
    *this *= p.inverse();
    return *this;
  }
  Mod_Int &operator++() { return *this += Mod_Int(1); }
  Mod_Int operator++(int) {
    Mod_Int tmp = *this;
    ++*this;
    return tmp;
  }
  Mod_Int &operator--() { return *this -= Mod_Int(1); }
  Mod_Int operator--(int) {
    Mod_Int tmp = *this;
    --*this;
    return tmp;
  }
  Mod_Int operator-() const { return Mod_Int(-x); }
  Mod_Int operator+(const Mod_Int &p) const { return Mod_Int(*this) += p; }
  Mod_Int operator-(const Mod_Int &p) const { return Mod_Int(*this) -= p; }
  Mod_Int operator*(const Mod_Int &p) const { return Mod_Int(*this) *= p; }
  Mod_Int operator/(const Mod_Int &p) const { return Mod_Int(*this) /= p; }
  bool operator==(const Mod_Int &p) const { return x == p.x; }
  bool operator!=(const Mod_Int &p) const { return x != p.x; }
  Mod_Int inverse() const {
    assert(*this != Mod_Int(0));
    return pow(mod - 2);
  }
  Mod_Int pow(long long k) const {
    Mod_Int now = *this, ret = 1;
    for (; k > 0; k >>= 1, now *= now) {
      if (k & 1) ret *= now;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const Mod_Int &p) {
    return os << p.x;
  }
  friend istream &operator>>(istream &is, Mod_Int &p) {
    long long a;
    is >> a;
    p = Mod_Int<mod>(a);
    return is;
  }
};
using mint = Mod_Int<MOD>;
int main() {
  int N;
  cin >> N;
  vector<mint> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  vector<vector<mint>> iv(N, vector<mint>(N));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      iv[i][j] = (a[i] + a[j]).inverse();
    }
  }
  vector<vector<mint>> dp1(N, vector<mint>(1 << N, 1));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 1 << N; j++) {
      for (int k = 0; k < N; k++) {
        if ((j >> k) & 1) {
          dp1[i][j] *= a[i] * iv[i][k];
        }
      }
    }
  }
  vector<mint> dp(1 << N, 1);
  for (int i = 0; i < 1 << N; i++) {
    for (int j = i; j > 0; j--, j &= i) {
      if (j == i) continue;
      mint tmp = 1;
      for (int k = 0; k < N; k++) {
        if ((j >> k) & 1) {
          tmp *= dp1[k][i & ~j];
        }
      }
      dp[i] -= dp[j] * tmp;
    }
  }
  mint ans = 0;
  for (int i = 0; i < 1 << N; i++) {
    if (i == 0) continue;
    int j = ((1 << N) - 1) ^ i;
    for (int k = 0; k < N; k++) {
      if ((i >> k) & 1) dp[i] *= dp1[k][j];
    }
    ans += dp[i] * __builtin_popcount(i);
  }
  cout << ans << '\n';
}

### Prompt

In CPP, your task is to solve 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;
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
template <typename T>
bool chmax(T &x, const T &y) {
  return (x < y) ? (x = y, true) : false;
};
template <typename T>
bool chmin(T &x, const T &y) {
  return (x > y) ? (x = y, true) : false;
};
struct io_setup {
  io_setup() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(15);
  }
} io_setup;
template <int mod>
struct Mod_Int {
  int x;
  Mod_Int() : x(0) {}
  Mod_Int(long long y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  static int get_mod() { return mod; }
  Mod_Int &operator+=(const Mod_Int &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  Mod_Int &operator-=(const Mod_Int &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  Mod_Int &operator*=(const Mod_Int &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  Mod_Int &operator/=(const Mod_Int &p) {
    *this *= p.inverse();
    return *this;
  }
  Mod_Int &operator++() { return *this += Mod_Int(1); }
  Mod_Int operator++(int) {
    Mod_Int tmp = *this;
    ++*this;
    return tmp;
  }
  Mod_Int &operator--() { return *this -= Mod_Int(1); }
  Mod_Int operator--(int) {
    Mod_Int tmp = *this;
    --*this;
    return tmp;
  }
  Mod_Int operator-() const { return Mod_Int(-x); }
  Mod_Int operator+(const Mod_Int &p) const { return Mod_Int(*this) += p; }
  Mod_Int operator-(const Mod_Int &p) const { return Mod_Int(*this) -= p; }
  Mod_Int operator*(const Mod_Int &p) const { return Mod_Int(*this) *= p; }
  Mod_Int operator/(const Mod_Int &p) const { return Mod_Int(*this) /= p; }
  bool operator==(const Mod_Int &p) const { return x == p.x; }
  bool operator!=(const Mod_Int &p) const { return x != p.x; }
  Mod_Int inverse() const {
    assert(*this != Mod_Int(0));
    return pow(mod - 2);
  }
  Mod_Int pow(long long k) const {
    Mod_Int now = *this, ret = 1;
    for (; k > 0; k >>= 1, now *= now) {
      if (k & 1) ret *= now;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const Mod_Int &p) {
    return os << p.x;
  }
  friend istream &operator>>(istream &is, Mod_Int &p) {
    long long a;
    is >> a;
    p = Mod_Int<mod>(a);
    return is;
  }
};
using mint = Mod_Int<MOD>;
int main() {
  int N;
  cin >> N;
  vector<mint> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  vector<vector<mint>> iv(N, vector<mint>(N));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      iv[i][j] = (a[i] + a[j]).inverse();
    }
  }
  vector<vector<mint>> dp1(N, vector<mint>(1 << N, 1));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 1 << N; j++) {
      for (int k = 0; k < N; k++) {
        if ((j >> k) & 1) {
          dp1[i][j] *= a[i] * iv[i][k];
        }
      }
    }
  }
  vector<mint> dp(1 << N, 1);
  for (int i = 0; i < 1 << N; i++) {
    for (int j = i; j > 0; j--, j &= i) {
      if (j == i) continue;
      mint tmp = 1;
      for (int k = 0; k < N; k++) {
        if ((j >> k) & 1) {
          tmp *= dp1[k][i & ~j];
        }
      }
      dp[i] -= dp[j] * tmp;
    }
  }
  mint ans = 0;
  for (int i = 0; i < 1 << N; i++) {
    if (i == 0) continue;
    int j = ((1 << N) - 1) ^ i;
    for (int k = 0; k < N; k++) {
      if ((i >> k) & 1) dp[i] *= dp1[k][j];
    }
    ans += dp[i] * __builtin_popcount(i);
  }
  cout << ans << '\n';
}

```