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;
  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;
  }
  