description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long int mult(long long int a, long long int b) {
    return ((a % int(1e9 + 7)) * (b % int(1e9 + 7))) % int(1e9 + 7);
  }
  long long int add(long long int a, long long int b) {
    return ((a % int(1e9 + 7)) + (b % int(1e9 + 7))) % int(1e9 + 7);
  }
  long long int subt(long long int a, long long int b) {
    long long int ans = ((a % int(1e9 + 7)) - (b % int(1e9 + 7))) % int(1e9 + 7);
    ans %= int(1e9 + 7);
    ans = (ans + int(1e9 + 7)) % int(1e9 + 7);
    return ans;
  }
  long long int exp(long long int a, long long int b) {
    long long int ans = 1;
    long long int po = a;
    while (b != 0) {
      if (b % 2) {
        ans = ((ans % int(1e9 + 7)) * (po % int(1e9 + 7))) % int(1e9 + 7);
      }
      po = ((po % int(1e9 + 7)) * (po % int(1e9 + 7))) % int(1e9 + 7);
      b /= 2;
    }
    return ans;
  }
  void nitheesh() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  signed main() {
    nitheesh();
    long long int t = 1;
    cin >> t;
    while (t--) {
      long long int n;
      cin >> n;
      for (long long i = 0; i < n; i++) {
        cout << i + 2 << " ";
      }
      cout << '\n';
    }
  }
  