human
I found an interesting problem on https://codeforces.com/problemset/problem/1423/K:
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 $gcd(a,b)$, $\frac{a}{gcd(a,b)}$, $\frac{b}{gcd(a,b)}$ 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 $t$ $(1 \leq t \leq 10^6)$ - number of test cases.

On next line there are $t$ numbers, $n_i$ $(1 \leq n_i \leq 10^6)$ - 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.

I tried it in Python, but could not do it. Can you solve it?

=====
Write me a Python program to print the answer on separate lines: number of lonely numbers in group $1, 2, 3,?
assistant
#Here is Python program to print the answer on separate lines: number of lonely numbers in group $1, 2, 3, .

import os
import sys
from io import BytesIO, IOBase

def main():
    pass

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

primes = []
prime = [True] * (10 ** 6  +5)
prime[0] = False

for i in range(2, 10 ** 6):
    if prime[i]:
        for j in range(2 * i, 10 ** 6 + 5, i):
            prime[j] = False

pref = [0]
for i in range(1, 10 ** 6 + 5):
    pref.append(pref[-1])
    if prime[i]:
        pref[-1] += 1
    s =...