#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1.01e9;
const long long infll = (long long)1.01e18;
const long double eps = 1e-9;
const long double pi = acos((long double)-1);
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x) { return mrand() % x; }
void precalc() {}
const int mod = (int)1e9 + 7;
int mul(int a, int b) { return (long long)a * b % mod; }
void add(int &a, int b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}
pair<int, int> gcd(int a, int b) {
  if (!b) {
    return make_pair(1, 0);
  }
  auto p = gcd(b, a % b);
  return make_pair(p.second, p.first - (a / b) * p.second);
}
int inv(int x) {
  int res = gcd(x, mod).first % mod;
  if (res < 0) {
    res += mod;
  }
  return res;
}
const int maxn = 15, pown = (1 << maxn);
int n;
int a[maxn];
bool read() {
  if (scanf("%d", &n) < 1) {
    return false;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  return true;
}
int g[maxn][maxn];
int mem[maxn][pown];
int dp[pown];
void solve() {
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      int q = inv(a[i] + a[j]);
      g[i][j] = mul(a[i], q);
      g[j][i] = mul(a[j], q);
    }
  }
  for (int v = 0; v < n; ++v) {
    for (int msk = 0; msk < (1 << n); ++msk) {
      auto &cur = mem[v][msk];
      cur = 1;
      for (int u = 0; u < n; ++u) {
        if (!(msk & (1 << u))) {
          continue;
        }
        cur = mul(cur, g[v][u]);
      }
    }
  }
  for (int msk = 1; msk < (1 << n); ++msk) {
    auto &cur = dp[msk];
    cur = 1;
    for (int sub = ((msk - 1) & msk); sub > 0; sub = ((sub - 1) & msk)) {
      int pr = dp[sub];
      for (int v = 0; v < n; ++v) {
        if (!(sub & (1 << v))) {
          continue;
        }
        pr = mul(pr, mem[v][msk ^ sub]);
      }
      add(cur, mod - pr);
    }
  }
  int res = 0;
  for (int msk = 1; msk < (1 << n); ++msk) {
    auto cur = dp[msk];
    for (int v = 0; v < n; ++v) {
      if (!(msk & (1 << v))) {
        continue;
      }
      cur = mul(cur, mem[v][((1 << n) - 1) ^ msk]);
    }
    add(res, mul(__builtin_popcount(msk), cur));
  }
  printf("%d\n", res);
}
int main() {
  precalc();
  while (read()) {
    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 inf = (int)1.01e9;
const long long infll = (long long)1.01e18;
const long double eps = 1e-9;
const long double pi = acos((long double)-1);
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x) { return mrand() % x; }
void precalc() {}
const int mod = (int)1e9 + 7;
int mul(int a, int b) { return (long long)a * b % mod; }
void add(int &a, int b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}
pair<int, int> gcd(int a, int b) {
  if (!b) {
    return make_pair(1, 0);
  }
  auto p = gcd(b, a % b);
  return make_pair(p.second, p.first - (a / b) * p.second);
}
int inv(int x) {
  int res = gcd(x, mod).first % mod;
  if (res < 0) {
    res += mod;
  }
  return res;
}
const int maxn = 15, pown = (1 << maxn);
int n;
int a[maxn];
bool read() {
  if (scanf("%d", &n) < 1) {
    return false;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  return true;
}
int g[maxn][maxn];
int mem[maxn][pown];
int dp[pown];
void solve() {
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      int q = inv(a[i] + a[j]);
      g[i][j] = mul(a[i], q);
      g[j][i] = mul(a[j], q);
    }
  }
  for (int v = 0; v < n; ++v) {
    for (int msk = 0; msk < (1 << n); ++msk) {
      auto &cur = mem[v][msk];
      cur = 1;
      for (int u = 0; u < n; ++u) {
        if (!(msk & (1 << u))) {
          continue;
        }
        cur = mul(cur, g[v][u]);
      }
    }
  }
  for (int msk = 1; msk < (1 << n); ++msk) {
    auto &cur = dp[msk];
    cur = 1;
    for (int sub = ((msk - 1) & msk); sub > 0; sub = ((sub - 1) & msk)) {
      int pr = dp[sub];
      for (int v = 0; v < n; ++v) {
        if (!(sub & (1 << v))) {
          continue;
        }
        pr = mul(pr, mem[v][msk ^ sub]);
      }
      add(cur, mod - pr);
    }
  }
  int res = 0;
  for (int msk = 1; msk < (1 << n); ++msk) {
    auto cur = dp[msk];
    for (int v = 0; v < n; ++v) {
      if (!(msk & (1 << v))) {
        continue;
      }
      cur = mul(cur, mem[v][((1 << n) - 1) ^ msk]);
    }
    add(res, mul(__builtin_popcount(msk), cur));
  }
  printf("%d\n", res);
}
int main() {
  precalc();
  while (read()) {
    solve();
  }
  return 0;
}

```