#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mminvprime(long long int a, long long int b) {
  return expo(a, b - 2, b);
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
vector<long long int> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  long long int start = 100005;
  for (long long int i = 0; i < n; i++) {
    a[i] = start;
    start = start + 1;
  }
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

Given n, find any array a_1, a_2, …, a_n of integers such that all of the following conditions hold: 

  * 1 ≤ a_i ≤ 10^9 for every i from 1 to n.

  * a_1 < a_2 < … <a_n

  * For every i from 2 to n, a_i isn't divisible by a_{i-1}




It can be shown that such an array always exists under the constraints of the problem.

Input

The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The only line of each test case contains a single integer n (1 ≤ n ≤ 1000).

It is guaranteed that the sum of n over all test cases does not exceed 10^4.

Output

For each test case print n integers a_1, a_2, …, a_n — the array you found. If there are multiple arrays satisfying all the conditions, print any of them.

Example

Input


3
1
2
7


Output


1
2 3
111 1111 11111 111111 1111111 11111111 111111111

Note

In the first test case, array [1] satisfies all the conditions.

In the second test case, array [2, 3] satisfies all the conditions, as 2<3 and 3 is not divisible by 2.

In the third test case, array [111, 1111, 11111, 111111, 1111111, 11111111, 111111111] satisfies all the conditions, as it's increasing and a_i isn't divisible by a_{i-1} for any i from 2 to 7.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mminvprime(long long int a, long long int b) {
  return expo(a, b - 2, b);
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
vector<long long int> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  long long int start = 100005;
  for (long long int i = 0; i < n; i++) {
    a[i] = start;
    start = start + 1;
  }
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```