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;
  template <typename T>
  class TreeNode {
   public:
    T data;
    std::vector<TreeNode<T>*> children;
    TreeNode(T data) { this->data = data; }
  };
  void print_tree(TreeNode<int>* root) {
    if (root == NULL) {
      return;
    }
    cout << root->data << ":";
    for (int i = 0; i < root->children.size(); i++) {
      cout << root->children[i]->data << ",";
    }
    cout << "\n";
    for (int i = 0; i < root->children.size(); i++) {
      print_tree(root->children[i]);
    }
  }
  int count_Nodes(TreeNode<int>* root) {
    int ans = 1;
    for (int i = 0; i < root->children.size(); i++) {
      ans += count_Nodes(root->children[i]);
    }
    return ans;
  }
  int max_Nodes(TreeNode<int>* root) {
    int ans = root->data;
    for (int i = 0; i < root->children.size(); i++) {
      ans = max(ans, max_Nodes(root->children[i]));
    }
    return ans;
  }
  int Sumof_Nodes(TreeNode<int>* root) {
    int ans = root->data;
    for (int i = 0; i < root->children.size(); i++) {
      ans += Sumof_Nodes(root->children[i]);
    }
    return ans;
  }
  int height_Nodes(TreeNode<int>* root) {
    int ans = 1;
    int max_ans = 0;
    for (int i = 0; i < root->children.size(); i++) {
      max_ans = max(max_ans, height_Nodes(root->children[i]));
    }
    ans += max_ans;
    return ans;
  }
  TreeNode<int>* Take_input() {
    int root_data;
    cin >> root_data;
    TreeNode<int>* root = new TreeNode<int>(root_data);
    std::queue<TreeNode<int>*> nodes;
    nodes.push(root);
    while (nodes.size() != 0) {
      TreeNode<int>* frontNode = nodes.front();
      nodes.pop();
      int num_child;
      cin >> num_child;
      for (int i = 0; i < num_child; i++) {
        int child_data;
        cin >> child_data;
        TreeNode<int>* child = new TreeNode<int>(child_data);
        frontNode->children.push_back(child);
        nodes.push(child);
      }
    }
    return root;
  }
  int f(int a, int b) {
    if (a - 2 < b) {
      return 1;
    } else {
      return f(9, 9);
    }
  }
  long long help[1000001];
  void helper() {
    for (long long i = 0; i < 1000001; i++) {
      help[i] = i;
    }
    for (long long i = 2; i < 1000001; i++) {
      for (long long j = 2 * i; j < 1000001; j += i) {
        if (j == help[j]) {
          help[j] = i;
        }
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    do {
      int n;
      cin >> n;
      vector<int> a(n);
      for (int i = 0; i < n; i++) {
        a[i] = i + 2;
      }
      for (auto x : a) {
        cout << x << " ";
      }
      cout << "\n";
    } while (--t);
    return 0;
  }
  