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 gcd(long long a, long long b) {
    if (a == 0) return b;
    gcd(b % a, a);
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  void input() {}
  int dx[] = {-1, 0, 0, 1, -1, -1, 1, 1};
  int dy[] = {0, 1, -1, 0, -1, 1, 1, -1};
  long long power(long long b, long long p) {
    long long an = 1;
    while (p > 0) {
      if (p % 2) {
        an *= b;
        an %= 1000000000;
      }
      b *= b;
      b %= 1000000000;
      p /= 2;
    }
    return an;
  }
  vector<bool> is_prime;
  void sieve(long long n) {
    is_prime.assign(n, true);
    is_prime[0] = is_prime[1] = 0;
    for (long long i = 2; i * i <= n; i++) {
      if (is_prime[i]) {
        for (long long j = i * i; j <= n; j += i) is_prime[j] = false;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin.tie(0);
    cin.sync_with_stdio(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 2; i < n + 2; i++) cout << i << " ";
      cout << '\n';
    }
    return 0;
  }
  