#include <bits/stdc++.h>
using namespace std;
long long ceilAB(long long a, long long b) { return (a + b - 1) / b; }
long long Pow(long long c, long long d) {
  return d == 0 ? 1 : c * pow(c, d - 1);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
const int mx = 1e6;
int a[mx];
void solve() {
  int t;
  cin >> t;
  while (t--) {
    int i, j, k, n, m, x, y, z, c = 0, cnt = 0;
    string s;
    cin >> n;
    for (i = 1; i <= n; i++) {
      cout << i + 1 << " ";
    }
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 ceilAB(long long a, long long b) { return (a + b - 1) / b; }
long long Pow(long long c, long long d) {
  return d == 0 ? 1 : c * pow(c, d - 1);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
const int mx = 1e6;
int a[mx];
void solve() {
  int t;
  cin >> t;
  while (t--) {
    int i, j, k, n, m, x, y, z, c = 0, cnt = 0;
    string s;
    cin >> n;
    for (i = 1; i <= n; i++) {
      cout << i + 1 << " ";
    }
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}

```