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 gcd(long long int a, long long int b) {
    return b ? gcd(b, a % b) : a;
  }
  long long int expo(long long int a, long long int b, long long int mod) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  long long int mminvprime(long long int a, long long int b) {
    return expo(a, b - 2, b);
  }
  long long int mod_add(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long int mod_mul(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long int mod_sub(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long int mod_div(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
  }
  vector<long long int> sieve(int n) {
    int* arr = new int[n + 1]();
    vector<long long int> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  void solve() {
    long long int n;
    cin >> n;
    long long int a[n];
    long long int start = 100005;
    for (long long int i = 0; i < n; i++) {
      a[i] = start;
      start = start + 1;
    }
    for (long long int i = 0; i < n; i++) {
      cout << a[i] << ' ';
    }
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  