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>
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("avx2")
  #pragma GCC optimize("Ofast")
  using namespace std;
  long long mod = 1000000007;
  template <typename T>
  void rv(vector<T> &a) {
    for (int i = 0; i < a.size(); i++) {
      cin >> a[i];
    }
  }
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  int lcm(int a, int b) { return a / gcd(a, b) * b; }
  uint64_t binpow(uint64_t a, uint64_t n) {
    uint64_t res = 1;
    while (n) {
      if (n & 1) res *= a;
      a *= a;
      n >>= 1;
    }
    return res;
  }
  long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  const int EPS = 9;
  const int eMXN = 2e5;
  long long MinDivisor(long long n) {
    if (n % 2 == 0) return 2;
    for (long long i = 3, isqrt = sqrt(n); i <= isqrt; i += 2)
      if (n % i == 0) return i;
    return n;
  }
  bool IsPrime(long long n) { return (MinDivisor(n) == n); }
  void print(vector<int> a) {
    for (auto c : a) {
      cout << c << " ";
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n = 10000;
      int x;
      cin >> x;
      int *a = new int[n + 1];
      int cnt = 0;
      for (int i = 0; i < n + 1; i++) a[i] = i;
      for (int p = 2; p < n + 1; p++) {
        if (a[p] != 0) {
          cout << a[p] << " ";
          cnt++;
          for (int j = p * p; j < n + 1; j += p) a[j] = 0;
        }
        if (cnt == x) {
          break;
        }
      }
      cout << "\n";
    }
    return 0;
  }
  