#include <bits/stdc++.h>
using namespace std;
long long isPrime[1000005];
long long pref[1000005];
void seive() {
  for (long long i = 2; i < 1000005; i++) {
    if (!isPrime[i])
      for (long long j = i + i; j < 1000005; j += i) isPrime[j] = 1;
    pref[i] = pref[i - 1] + 1 - isPrime[i];
  }
}
int main() {
  seive();
  long long t;
  scanf("%lld", &t);
  while (t-- > 0) {
    long long n;
    scanf("%lld", &n);
    long long x = sqrt(n);
    long long ans = pref[n] - pref[x] + 1;
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long isPrime[1000005];
long long pref[1000005];
void seive() {
  for (long long i = 2; i < 1000005; i++) {
    if (!isPrime[i])
      for (long long j = i + i; j < 1000005; j += i) isPrime[j] = 1;
    pref[i] = pref[i - 1] + 1 - isPrime[i];
  }
}
int main() {
  seive();
  long long t;
  scanf("%lld", &t);
  while (t-- > 0) {
    long long n;
    scanf("%lld", &n);
    long long x = sqrt(n);
    long long ans = pref[n] - pref[x] + 1;
    printf("%lld\n", ans);
  }
  return 0;
}

```