#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
long long sum(long long n) { return (n * (n + 1)) / 2.0; }
long long len(string str) { return str.length(); }
long long fact(long long n) {
  if (n > 0)
    return n * fact(n - 1);
  else
    return 1;
}
void solve(void);
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, b % a);
}
bool sortbysec(const pair<string, int> &a, const pair<string, int> &b) {
  return (a.second > b.second);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cout << (i + 2) << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test = 1, i = 1;
  cin >> test;
  while (test--) {
    solve();
    cout << '\n';
  }
}

### Prompt

Please create a solution in Cpp 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;
const long long MOD = 1000000007;
long long sum(long long n) { return (n * (n + 1)) / 2.0; }
long long len(string str) { return str.length(); }
long long fact(long long n) {
  if (n > 0)
    return n * fact(n - 1);
  else
    return 1;
}
void solve(void);
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, b % a);
}
bool sortbysec(const pair<string, int> &a, const pair<string, int> &b) {
  return (a.second > b.second);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cout << (i + 2) << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test = 1, i = 1;
  cin >> test;
  while (test--) {
    solve();
    cout << '\n';
  }
}

```