#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793;
const int Mod[] = {998244353, (int)1e9 + 7, 1223456789, 1234567891, 2147483647};
const int mod = Mod[1];
const int N = 2e6 + 5;
const int NN = 1e7 + 5;
const int M = 1e3 + 5;
const int MM = 1e4;
const long long oo = 9223372036854775807ll;
typedef bool (*func)(int &, int);
typedef bool (*funcl)(long long &, long long);
template <class t>
bool add(t &x, t y) {
  x += y;
  if (x >= mod) {
    x -= mod;
    return 0;
  }
  return 1;
}
template <class t>
bool mius(t &x, t y) {
  x -= y;
  if (x < 0) {
    x += mod;
    return 0;
  }
  return 1;
}
template <class t>
bool mini(t &x, t y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <class t>
bool maxi(t &x, t y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
template <class t>
bool doxor(t &x, t y) {
  x ^= y;
  return (x > 0);
}
template <class T>
class DSU {
 private:
  T n;
  vector<T> cha;

 public:
  DSU(const T &n = 0) : n(n + 7), cha(n + 7, -1) {}
  T root(T u) { return (cha[u] < 0 ? u : cha[u] = root(cha[u])); }
  bool join(T u, T v) {
    if ((v = root(v)) == (u = root(u))) return 0;
    if (cha[u] > cha[v]) swap(u, v);
    cha[u] += cha[v];
    cha[v] = u;
    return 1;
  }
};
template <class T, class TT, class up>
class FenWick {
 private:
  T n;
  TT init;
  vector<TT> bit;
  up update;

 public:
  FenWick(const T &n = 0, const up &update = nullptr, const TT &init = 0)
      : n(n + 7), init(init), bit(n + 7, init), update(update) {}
  void adj(T x, TT val) {
    for (; x <= n; x += x & -x) update(bit[x], val);
  }
  TT get(T x) {
    TT res = init;
    for (; x > 0; x -= x & -x) update(res, bit[x]);
    return res;
  }
};
void file(string name = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  return;
}
long long gt[N], rgt[N];
long long power(long long a, long long b) {
  if (b == 0) return 1ll;
  if (b == 1) return a;
  long long tam = 1ll * power(a, b / 2);
  tam = 1ll * tam * tam % mod;
  if (b & 1) return 1ll * tam * a % mod;
  return tam;
}
void prepare() {
  gt[0] = 1;
  for (int i = 1; i <= N; i++) gt[i] = 1ll * gt[i - 1] * i % mod;
}
int nCk(int n, int k) {
  return 1ll * gt[n] * power(gt[n - k], mod - 2) % mod * power(gt[k], mod - 2) %
         mod;
}
void solve(int test = -1) {
  int n;
  cin >> n;
  long long res = 1;
  for (int i = 1; i <= n; i++) {
    res *= n * 2 - i;
    res %= mod;
  }
  res *= gt[n];
  res %= mod;
  cout << res;
  return;
}
int32_t main() {
  file();
  int T = 1;
  prepare();
  cin >> T;
  for (int i = 1; i <= T; i++) {
    solve(i);
    cout << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
const double PI = 3.141592653589793;
const int Mod[] = {998244353, (int)1e9 + 7, 1223456789, 1234567891, 2147483647};
const int mod = Mod[1];
const int N = 2e6 + 5;
const int NN = 1e7 + 5;
const int M = 1e3 + 5;
const int MM = 1e4;
const long long oo = 9223372036854775807ll;
typedef bool (*func)(int &, int);
typedef bool (*funcl)(long long &, long long);
template <class t>
bool add(t &x, t y) {
  x += y;
  if (x >= mod) {
    x -= mod;
    return 0;
  }
  return 1;
}
template <class t>
bool mius(t &x, t y) {
  x -= y;
  if (x < 0) {
    x += mod;
    return 0;
  }
  return 1;
}
template <class t>
bool mini(t &x, t y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <class t>
bool maxi(t &x, t y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
template <class t>
bool doxor(t &x, t y) {
  x ^= y;
  return (x > 0);
}
template <class T>
class DSU {
 private:
  T n;
  vector<T> cha;

 public:
  DSU(const T &n = 0) : n(n + 7), cha(n + 7, -1) {}
  T root(T u) { return (cha[u] < 0 ? u : cha[u] = root(cha[u])); }
  bool join(T u, T v) {
    if ((v = root(v)) == (u = root(u))) return 0;
    if (cha[u] > cha[v]) swap(u, v);
    cha[u] += cha[v];
    cha[v] = u;
    return 1;
  }
};
template <class T, class TT, class up>
class FenWick {
 private:
  T n;
  TT init;
  vector<TT> bit;
  up update;

 public:
  FenWick(const T &n = 0, const up &update = nullptr, const TT &init = 0)
      : n(n + 7), init(init), bit(n + 7, init), update(update) {}
  void adj(T x, TT val) {
    for (; x <= n; x += x & -x) update(bit[x], val);
  }
  TT get(T x) {
    TT res = init;
    for (; x > 0; x -= x & -x) update(res, bit[x]);
    return res;
  }
};
void file(string name = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  return;
}
long long gt[N], rgt[N];
long long power(long long a, long long b) {
  if (b == 0) return 1ll;
  if (b == 1) return a;
  long long tam = 1ll * power(a, b / 2);
  tam = 1ll * tam * tam % mod;
  if (b & 1) return 1ll * tam * a % mod;
  return tam;
}
void prepare() {
  gt[0] = 1;
  for (int i = 1; i <= N; i++) gt[i] = 1ll * gt[i - 1] * i % mod;
}
int nCk(int n, int k) {
  return 1ll * gt[n] * power(gt[n - k], mod - 2) % mod * power(gt[k], mod - 2) %
         mod;
}
void solve(int test = -1) {
  int n;
  cin >> n;
  long long res = 1;
  for (int i = 1; i <= n; i++) {
    res *= n * 2 - i;
    res %= mod;
  }
  res *= gt[n];
  res %= mod;
  cout << res;
  return;
}
int32_t main() {
  file();
  int T = 1;
  prepare();
  cin >> T;
  for (int i = 1; i <= T; i++) {
    solve(i);
    cout << '\n';
  }
  return 0;
}

```