Iterate

a, b = input().split()
n = int(a + b)

flag = False
for i in range(n):
    if i ** 2 == n:
        flag = True
        break

print('Yes' if flag else 'No')


# Why is this code slow?

# This code is slow because it is using a brute force approach to find the square root of the input number. It is looping through every possible number starting from 0 until n. Note that the sqare root will be smaller than n, so at least half of the numbers it is looping through are unnecessary. At most, you need to loop through the numbers up to the square root of n.

# Improved version:

a, b = input().split()
n = int(a + b)

flag = False
i = 0
while i * i <= n:
    if i * i == n:
        flag = True
        break
    i += 1

print('Yes' if flag else 'No')

### END ###

def main():
    N, *A = map(int, open(0).read().split())
    remaining = 0
    cur = 0  # current index
    for target in range(1, N + 1):
        be_kept = False
        for i in range(cur, N):
            if A[i] == target:
                remaining += 1
                cur = i + 1
                be_kept = True
                break
        if not be_kept:
            break
    print(N - remaining if remaining else -1)

# Why is this code slow?

# This code is slow because it is using a brute force approach to search for the target number in the list of numbers. It is looping through the list for every target number, which can take a long time if the list is very large. A more efficient approach would be to use a data structure such as a hash table, which can perform lookups in constant time.

# Improved version:

def main():
    N, *A = map(int, open(0).read().split())
    remaining = 0
    target = 1
    for i in A:
        if i == target:
            remaining += 1
            target += 1
    print(N - remaining if remaining else -1)


if __name__ == '__main__':
    main()


### END ###

import numpy as np

N = int(input())
n=int(np.sqrt(N))
print(n**2)

# Why is this code slow?

# This code is slow because it is using numpy for calculating the square root of the input number. Numpy is much slower than using the built-in math module, which can calculate the square root faster.

# Improved version:

N = int(input())
n = int(N**0.5) 
print(n**2)

### END ###

import numpy as np
A = np.arange(1<<27,dtype=np.int32)


a,b = map(int,input().split())
if (a-b) % 2 == 0:
  print((a+b)//2)
else:
  print('IMPOSSIBLE')

# Why is this code slow?

# This code is slow because it is using numpy to calculate the range of numbers from 1 to 2^27. This is an extremely large range and numpy is slow at calculating such a large range. A more efficient approach would be to use a loop to calculate the range, which would be much faster.

# Improved version:

import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

A,B = map(int,read().split())

q,r = divmod(A+B,2)

if r == 1:
    print('IMPOSSIBLE')
else:
    print(q)

### END ###