#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
long long int Pow(long long int c, long long int d) {
  return d == 0 ? 1 : c * pow(c, d - 1);
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return ((a * b) / gcd(a, b));
}
long long int egcd(long long int a, long long int b, long long int &x,
                   long long int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int gcd = egcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return gcd;
}
long long int bigMod(long long int a, long long int b) {
  a %= mod;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
inline long long int MOD(long long int a) { return (a % mod + mod) % mod; }
inline long long int modAdd(long long int a, long long int b) {
  return MOD(MOD(a) + MOD(b));
}
inline long long int modSub(long long int a, long long int b) {
  return MOD(MOD(a) - MOD(b));
}
inline long long int modMul(long long int a, long long int b) {
  return MOD(MOD(a) * MOD(b));
}
inline long long int modInv(long long int a) { return bigMod(a, mod - 2); }
inline long long int modDiv(long long int a, long long int b) {
  a = MOD(a);
  b = modInv(b);
  return (a * b) % mod;
}
vector<string> v;
map<char, long long int> mp;
bool compare(pair<long long int, pair<long long int, long long int>> p1,
             pair<long long int, pair<long long int, long long int>> p2) {
  if (p1.first == p2.first and p1.second.second > p2.second.second)
    return false;
  else if (p1.first < p2.first)
    return false;
  else
    return true;
}
set<string> vs;
void permute(string str, string out) {
  if (str.size() == 0) {
    vs.insert(out);
    return;
  }
  for (int i = 0; i < str.size(); i++) {
    permute(str.substr(1), out + str[0]);
    rotate(str.begin(), str.begin() + 1, str.end());
  }
}
bool isPrime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int ncr(long long int nn, long long int rr) {
  if (rr > nn) return 0;
  if (rr == 0) return 1;
  if (rr == 1) return nn;
  if (nn == rr) return 1;
  return ncr(nn - 1, rr - 1) + ncr(nn - 1, rr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  long long int fact[200001];
  fact[0] = 1;
  for (long long int i = 1; i <= 200000; i++) fact[i] = modMul(fact[i - 1], i);
  while (t--) {
    long long int n;
    cin >> n;
    n = n * 2;
    long long int ans = modDiv(fact[n], 2);
    cout << ans << endl;
  }
  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 long long int mod = 1e9 + 7;
long long int Pow(long long int c, long long int d) {
  return d == 0 ? 1 : c * pow(c, d - 1);
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return ((a * b) / gcd(a, b));
}
long long int egcd(long long int a, long long int b, long long int &x,
                   long long int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int gcd = egcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return gcd;
}
long long int bigMod(long long int a, long long int b) {
  a %= mod;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
inline long long int MOD(long long int a) { return (a % mod + mod) % mod; }
inline long long int modAdd(long long int a, long long int b) {
  return MOD(MOD(a) + MOD(b));
}
inline long long int modSub(long long int a, long long int b) {
  return MOD(MOD(a) - MOD(b));
}
inline long long int modMul(long long int a, long long int b) {
  return MOD(MOD(a) * MOD(b));
}
inline long long int modInv(long long int a) { return bigMod(a, mod - 2); }
inline long long int modDiv(long long int a, long long int b) {
  a = MOD(a);
  b = modInv(b);
  return (a * b) % mod;
}
vector<string> v;
map<char, long long int> mp;
bool compare(pair<long long int, pair<long long int, long long int>> p1,
             pair<long long int, pair<long long int, long long int>> p2) {
  if (p1.first == p2.first and p1.second.second > p2.second.second)
    return false;
  else if (p1.first < p2.first)
    return false;
  else
    return true;
}
set<string> vs;
void permute(string str, string out) {
  if (str.size() == 0) {
    vs.insert(out);
    return;
  }
  for (int i = 0; i < str.size(); i++) {
    permute(str.substr(1), out + str[0]);
    rotate(str.begin(), str.begin() + 1, str.end());
  }
}
bool isPrime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int ncr(long long int nn, long long int rr) {
  if (rr > nn) return 0;
  if (rr == 0) return 1;
  if (rr == 1) return nn;
  if (nn == rr) return 1;
  return ncr(nn - 1, rr - 1) + ncr(nn - 1, rr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  long long int fact[200001];
  fact[0] = 1;
  for (long long int i = 1; i <= 200000; i++) fact[i] = modMul(fact[i - 1], i);
  while (t--) {
    long long int n;
    cin >> n;
    n = n * 2;
    long long int ans = modDiv(fact[n], 2);
    cout << ans << endl;
  }
  return 0;
}

```