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;
  vector<bool> prime(1000001, true);
  long long cha;
  const int INF = 1000000000;
  vector<vector<int>> g(0);
  vector<pair<long long, long long>> t(0);
  pair<int, int> combine(pair<int, int> a, pair<int, int> b) {
    if (a.first > b.first) return a;
    if (b.first > a.first) return b;
    return make_pair(a.first, a.second + b.second);
  }
  void build(vector<long long> a, int v, int tl, int tr) {
    if (tl == tr)
      t[v] = make_pair(a[tl], 1);
    else {
      int tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = combine(t[v * 2], t[v * 2 + 1]);
    }
  }
  pair<int, int> get_max(int v, int tl, int tr, int l, int r) {
    if (l > r) return make_pair(-INF, 0);
    if (l == tl && r == tr) return t[v];
    int tm = (tl + tr) / 2;
    return combine(get_max(v * 2, tl, tm, l, min(r, tm)),
                   get_max(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r));
  }
  void update(int v, int tl, int tr, int pos, int new_val) {
    if (tl == tr)
      t[v] = make_pair(new_val, 1);
    else {
      int tm = (tl + tr) / 2;
      if (pos <= tm)
        update(v * 2, tl, tm, pos, new_val);
      else
        update(v * 2 + 1, tm + 1, tr, pos, new_val);
      t[v] = combine(t[v * 2], t[v * 2 + 1]);
    }
  }
  void swap(long long a, long long b) {
    a += b;
    b = a - b;
    a -= b;
  }
  unsigned long long binpow(long long a, long long n) {
    int res = 1;
    while (n) {
      if (n & 1) res *= a;
      a *= a;
      n >>= 1;
    }
    return res;
  }
  int phi(int n) {
    int result = n;
    for (int i = 2; i * i <= n; ++i)
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        result -= result / i;
      }
    if (n > 1) result -= result / n;
    return result;
  }
  void pr(int n) {
    prime[0] = prime[1] = false;
    for (int i = 2; i <= n; ++i)
      if (prime[i])
        if (i * 1ll * i <= n)
          for (int j = i * i; j <= n; j += i) prime[j] = false;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) {
      if (a == 0) {
        return -1;
      }
      return a;
    }
    return gcd(abs(b - a), a);
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tt, n;
    cin >> tt;
    pr(100000);
    vector<int> pri(1000);
    int last = 0;
    for (int i = 0; i < 1000; i++) {
      while (!prime[last]) {
        last++;
      }
      pri[i] = last;
      last++;
    }
    while (tt--) {
      cin >> n;
      for (int i = 0; i < n; i++) {
        cout << pri[i] << " ";
      }
      cout << endl;
    }
  }
  