#include <bits/stdc++.h>
using namespace std;
void _print(long long int t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first != b.first)
    return a.first > b.first;
  else
    return a.second > b.second;
}
long long int binarySearch(vector<long long int> v, long long int l,
                           long long int r, long long int x) {
  while (l <= r) {
    long long int mid = l + (r - l) / 2;
    if (v[mid] == x)
      return mid;
    else if (v[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
long long int power(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return power(x * x, n / 2);
  else
    return x * power(x * x, (n - 1) / 2);
}
long long int ModularExpo(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) cout << i + 2 << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
void _print(long long int t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first != b.first)
    return a.first > b.first;
  else
    return a.second > b.second;
}
long long int binarySearch(vector<long long int> v, long long int l,
                           long long int r, long long int x) {
  while (l <= r) {
    long long int mid = l + (r - l) / 2;
    if (v[mid] == x)
      return mid;
    else if (v[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
long long int power(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return power(x * x, n / 2);
  else
    return x * power(x * x, (n - 1) / 2);
}
long long int ModularExpo(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++) cout << i + 2 << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```