#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long long int inf = 9 * 1e18;
const long long maxn = 2e5 + 100;
const int MAXN = 1.1e6;
int spf[MAXN];
void sieve() {
  for (int i = 1; i < MAXN; i++) spf[i] = 1;
  for (int i = 2; i * i < MAXN; i++) {
    for (int j = i * i; j < MAXN; j += i * i) spf[j] = i;
  }
}
long long fact[MAXN], inv[MAXN], invFact[MAXN];
void factGen() {
  fact[0] = inv[1] = fact[1] = invFact[0] = invFact[1] = 1;
  for (long long i = 2; i < MAXN; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
    inv[i] = mod - (inv[mod % i] * (mod / i) % mod);
    invFact[i] = (inv[i] * invFact[i - 1]) % mod;
  }
}
int gcdExtended(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverse(int a, int m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    int res = (x % m + m) % m;
    return res;
  }
}
long long binpow(long long a, long long b, long long m) {
  if (b < 0) return 0LL;
  if (a <= 0) return 0LL;
  a %= m;
  long long ans = 1LL;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long modinv(long long n) { return binpow(n, mod - 2, mod); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void solve() {
  int n;
  cin >> n;
  cout << (fact[2 * n - 1] * n) % mod << "\n";
}
int main() {
  factGen();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
long long int mod = 1e9 + 7;
long long int inf = 9 * 1e18;
const long long maxn = 2e5 + 100;
const int MAXN = 1.1e6;
int spf[MAXN];
void sieve() {
  for (int i = 1; i < MAXN; i++) spf[i] = 1;
  for (int i = 2; i * i < MAXN; i++) {
    for (int j = i * i; j < MAXN; j += i * i) spf[j] = i;
  }
}
long long fact[MAXN], inv[MAXN], invFact[MAXN];
void factGen() {
  fact[0] = inv[1] = fact[1] = invFact[0] = invFact[1] = 1;
  for (long long i = 2; i < MAXN; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
    inv[i] = mod - (inv[mod % i] * (mod / i) % mod);
    invFact[i] = (inv[i] * invFact[i - 1]) % mod;
  }
}
int gcdExtended(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverse(int a, int m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    int res = (x % m + m) % m;
    return res;
  }
}
long long binpow(long long a, long long b, long long m) {
  if (b < 0) return 0LL;
  if (a <= 0) return 0LL;
  a %= m;
  long long ans = 1LL;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long modinv(long long n) { return binpow(n, mod - 2, mod); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void solve() {
  int n;
  cin >> n;
  cout << (fact[2 * n - 1] * n) % mod << "\n";
}
int main() {
  factGen();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```