#include <bits/stdc++.h>
const int N = 100001, M = 998244353, mod = 1000000007;
const long long MX = INT64_MAX, MN = INT64_MIN, oo = 1e18;
using namespace std;
vector<string> tokenizer(string str, char ch) {
  std::istringstream var((str));
  vector<string> v;
  string t;
  while (getline((var), t, (ch))) {
    v.emplace_back(t);
  }
  return v;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void put(T&&... args) {
  ((cout << args << " "), ...);
}
void file_i_o() {
  freopen("./tests/test01.txt", "r", stdin);
  freopen("./tests/output01.txt", "w", stdout);
}
long long expm(long long a, long long b, long long m = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = ((res % m) * (a % m)) % m;
    }
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return res;
}
void solve(int T) {
  long long n;
  read(n);
  long long ans = 1;
  for (int i = 2; i < 2 * n + 1; i++) ans = (ans * i) % mod;
  {
    put((ans * expm(2, mod - 2)) % mod);
    cout << "\n";
  }
}
int main(int argc, char const* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 1; i < t + 1; i++) {
    solve(t);
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

CQXYM is counting permutations length of 2n.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

A permutation p(length of 2n) will be counted only if the number of i satisfying p_i<p_{i+1} is no less than n. For example:

  * Permutation [1, 2, 3, 4] will count, because the number of such i that p_i<p_{i+1} equals 3 (i = 1, i = 2, i = 3).
  * Permutation [3, 2, 1, 4] won't count, because the number of such i that p_i<p_{i+1} equals 1 (i = 3). 



CQXYM wants you to help him to count the number of such permutations modulo 1000000007 (10^9+7).

In addition, [modulo operation](https://en.wikipedia.org/wiki/Modulo_operation) is to get the remainder. For example:

  * 7 mod 3=1, because 7 = 3 ⋅ 2 + 1, 
  * 15 mod 4=3, because 15 = 4 ⋅ 3 + 3. 

Input

The input consists of multiple test cases. 

The first line contains an integer t (t ≥ 1) — the number of test cases. The description of the test cases follows.

Only one line of each test case contains an integer n(1 ≤ n ≤ 10^5).

It is guaranteed that the sum of n over all test cases does not exceed 10^5

Output

For each test case, print the answer in a single line.

Example

Input


4
1
2
9
91234


Output


1
12
830455698
890287984

Note

n=1, there is only one permutation that satisfies the condition: [1,2].

In permutation [1,2], p_1<p_2, and there is one i=1 satisfy the condition. Since 1 ≥ n, this permutation should be counted. In permutation [2,1], p_1>p_2. Because 0<n, this permutation should not be counted.

n=2, there are 12 permutations: [1,2,3,4],[1,2,4,3],[1,3,2,4],[1,3,4,2],[1,4,2,3],[2,1,3,4],[2,3,1,4],[2,3,4,1],[2,4,1,3],[3,1,2,4],[3,4,1,2],[4,1,2,3].

### Response

```cpp
#include <bits/stdc++.h>
const int N = 100001, M = 998244353, mod = 1000000007;
const long long MX = INT64_MAX, MN = INT64_MIN, oo = 1e18;
using namespace std;
vector<string> tokenizer(string str, char ch) {
  std::istringstream var((str));
  vector<string> v;
  string t;
  while (getline((var), t, (ch))) {
    v.emplace_back(t);
  }
  return v;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void put(T&&... args) {
  ((cout << args << " "), ...);
}
void file_i_o() {
  freopen("./tests/test01.txt", "r", stdin);
  freopen("./tests/output01.txt", "w", stdout);
}
long long expm(long long a, long long b, long long m = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = ((res % m) * (a % m)) % m;
    }
    b /= 2;
    a = ((a % m) * (a % m)) % m;
  }
  return res;
}
void solve(int T) {
  long long n;
  read(n);
  long long ans = 1;
  for (int i = 2; i < 2 * n + 1; i++) ans = (ans * i) % mod;
  {
    put((ans * expm(2, mod - 2)) % mod);
    cout << "\n";
  }
}
int main(int argc, char const* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 1; i < t + 1; i++) {
    solve(t);
  }
  return 0;
}

```