#include <bits/stdc++.h>
using lli = long long int;
using li = long int;
using namespace std;
const lli MOD{1000000007};
const lli INF{(false ? (INT_MAX - 1) / 2 : (lli)1e18 + 5)};
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  const int N{(int)1e6 + 1};
  vector<bool> isPrime(N, true);
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i{2}; (lli)i * i <= N; ++i) {
    if (isPrime[i])
      for (int j{i * i}; j <= N; j += i) isPrime[j] = false;
  }
  vector<int> p(N);
  p[0] = 1;
  for (int i{1}; i < N; ++i) p[i] = p[i - 1] + isPrime[i];
  li T{1};
  cin >> T;
  while (T--) {
    int n{};
    cin >> n;
    int s{(int)sqrt(n)};
    cout << p[n] - p[s] + 1 << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.

More precisely, two different numbers a and b are friends if gcd(a,b), (a)/(gcd(a,b)), (b)/(gcd(a,b)) can form sides of a triangle.

Three numbers a, b and c can form sides of a triangle if a + b > c, b + c > a and c + a > b.

In a group of numbers, a number is lonely if it doesn't have any friends in that group.

Given a group of numbers containing all numbers from 1, 2, 3, ..., n, how many numbers in that group are lonely?

Input

The first line contains a single integer t (1 ≤ t ≤ 10^6) - number of test cases.

On next line there are t numbers, n_i (1 ≤ n_i ≤ 10^6) - meaning that in case i you should solve for numbers 1, 2, 3, ..., n_i.

Output

For each test case, print the answer on separate lines: number of lonely numbers in group 1, 2, 3, ..., n_i.

Example

Input


3
1 5 10


Output


1
3
3

Note

For first test case, 1 is the only number and therefore lonely.

For second test case where n=5, numbers 1, 3 and 5 are lonely.

For third test case where n=10, numbers 1, 5 and 7 are lonely.

### Response

```cpp
#include <bits/stdc++.h>
using lli = long long int;
using li = long int;
using namespace std;
const lli MOD{1000000007};
const lli INF{(false ? (INT_MAX - 1) / 2 : (lli)1e18 + 5)};
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  const int N{(int)1e6 + 1};
  vector<bool> isPrime(N, true);
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i{2}; (lli)i * i <= N; ++i) {
    if (isPrime[i])
      for (int j{i * i}; j <= N; j += i) isPrime[j] = false;
  }
  vector<int> p(N);
  p[0] = 1;
  for (int i{1}; i < N; ++i) p[i] = p[i - 1] + isPrime[i];
  li T{1};
  cin >> T;
  while (T--) {
    int n{};
    cin >> n;
    int s{(int)sqrt(n)};
    cout << p[n] - p[s] + 1 << '\n';
  }
  return 0;
}

```