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;
  inline void wrt(int x, const char endc = 10) {
    static char buf[21], *nw;
    nw = buf;
    if (!x) {
      putchar(48);
      goto end;
    }
    if (x < 0) putchar(45), x = -x;
    while (x) *(nw++) = x % 10 + 48, x /= 10;
    while (nw != buf) putchar(*(--nw));
  end:
    putchar(endc);
  }
  inline int _isdigit(const char x) { return x > 47 and x < 58; }
  inline int rd() {
    static int f, x;
    static char c;
    f = x = 0;
    do
      if ((c = getchar()) == 45) f = 1;
    while (!_isdigit(c));
    do x = (x << 1) + (x << 3) + c - 48;
    while (_isdigit(c = getchar()));
    return f ? -x : x;
  }
  void work() {
    for (int i = 1, iend = rd(); i <= iend; i++) wrt(i + 1, 32);
    puts("");
  }
  int main() {
    for (int T = 1, Tend = rd(); T <= Tend; T++) work();
  }
  