#include <bits/stdc++.h>
using namespace std;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long idx,
               __attribute__((unused)) long long LINE_NUM) {
  cerr << '\n';
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long idx, long long LINE_NUM, Head H,
               Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
void run() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
}
const long long N = 2e5 + 5, MOD = 1e9 + 7, mod2 = 998244353, MAXN = 1e6 + 29;
void files() {}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cout << i + 2 << ' ';
  }
  cout << '\n';
}
signed main() {
  files();
  run();
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long idx,
               __attribute__((unused)) long long LINE_NUM) {
  cerr << '\n';
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long idx, long long LINE_NUM, Head H,
               Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
void run() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
}
const long long N = 2e5 + 5, MOD = 1e9 + 7, mod2 = 998244353, MAXN = 1e6 + 29;
void files() {}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cout << i + 2 << ' ';
  }
  cout << '\n';
}
signed main() {
  files();
  run();
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```