#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7, M = 2 * N;
int ans[N], sub[N];
const int NN = 1e6 + 7;
int comp[NN], primes[NN], cnt;
void sieve() {
  cnt = 0;
  int op = 0;
  comp[1] = 0;
  for (int i = 2; i <= NN; ++i) {
    int &x = comp[i];
    if (!x) primes[cnt++] = x = i, op++;
    for (int j = 0; primes[j] <= NN / i; j++) {
      op++;
      comp[i * primes[j]] = primes[j];
      if (primes[j] >= x) break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  sieve();
  ans[1] = 1;
  for (int i = 2; i < N; ++i) {
    ans[i] = ans[i - 1] - sub[i];
    if (comp[i] == i) {
      ans[i]++;
      if (i <= 1e3) sub[i * i]++;
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << ans[n] << '\n';
  }
}

### Prompt

Your challenge is to write a Cpp solution 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 namespace std;
const int N = 1e6 + 7, M = 2 * N;
int ans[N], sub[N];
const int NN = 1e6 + 7;
int comp[NN], primes[NN], cnt;
void sieve() {
  cnt = 0;
  int op = 0;
  comp[1] = 0;
  for (int i = 2; i <= NN; ++i) {
    int &x = comp[i];
    if (!x) primes[cnt++] = x = i, op++;
    for (int j = 0; primes[j] <= NN / i; j++) {
      op++;
      comp[i * primes[j]] = primes[j];
      if (primes[j] >= x) break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  sieve();
  ans[1] = 1;
  for (int i = 2; i < N; ++i) {
    ans[i] = ans[i - 1] - sub[i];
    if (comp[i] == i) {
      ans[i]++;
      if (i <= 1e3) sub[i * i]++;
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << ans[n] << '\n';
  }
}

```