Given a number n. Find the last two digits of 5 ^ n ( 5 to the power of n ).
Remember that overflow can occur.

-----Input:-----
- N  — the power in which you need to raise number 5.

-----Output:-----
Last two digits of 5^n.

-----Constraints-----
- $2 \leq N \leq 2.1018$

-----Sample Input:-----
2

-----Sample Output:-----
25

Difficulty: interview

Starter code:

Solutions:
print(25)

import sys
import math
import bisect
from sys import stdin,stdout
from math import gcd,floor,sqrt,log
from collections import defaultdict as dd
from bisect import bisect_left as bl,bisect_right as br

sys.setrecursionlimit(100000000)

ii     =lambda: int(input())
si     =lambda: input()
jn     =lambda x,l: x.join(map(str,l))
sl     =lambda: list(map(str,input().strip()))
mi     =lambda: list(map(int,input().split()))
mif    =lambda: list(map(float,input().split()))
lii    =lambda: list(map(int,input().split()))

ceil   =lambda x: int(x) if(x==int(x)) else int(x)+1
ceildiv=lambda x,d: x//d if(x%d==0) else x//d+1

flush  =lambda: stdout.flush()
stdstr =lambda: stdin.readline()
stdint =lambda: int(stdin.readline())
stdpr  =lambda x: stdout.write(str(x))

mod=1000000007


#main code
n=ii()
print(25)



# cook your dish here
x = int(input())

if(x==0):
    print(0)
elif(x==1):
    print(5)
else:
    print(25)

_ = input()
print(25)

N=float(eval(input()))
print(25)


# cook your dish here
n=float(input())
print(25)

# cook your dish here
def power(x, y, p):
    res = 1
    x = x % p
    while (y > 0):
        if (y & 1):
            res = (res * x) % p 
        y = y >> 1 
        x = (x * x) % p 
          
    return res 

def numberOfDigits(x): 
  
    i = 0
    while (x): 
        x //= 10
        i += 1
      
    return i 

def LastTwoDigit(n): 
  
    temp = 1
    for i in range(1, 3): 
        temp *= 10
  
    temp = power(5, n, temp) 

    for i in range(2 - numberOfDigits(temp)): 
        print(0, end = "") 
  
    if temp: 
        print(temp) 
  
def __starting_point(): 
    n = int(input())
    LastTwoDigit(n)
    
__starting_point()

print(25)

print(25)

def p(x,y,p):
    res=1
    x=x%p
    while(y>0):
        if(y&1):
            res=(res*x)%p
        y=y>>1
        x=(x*x)%p
    return res
def nod(x):
    i=0
    while(x):
        x//=10
        i+=1
    return i
def ltd(n):
    temp=1
    for i in range(1,3):
        temp*=10
        temp=p(5,n,temp)
    for i in range(2-nod(temp)):
        print(0,end="")
    if temp:
        print(temp)
def __starting_point():
    n=int(input())
    ltd(n)
__starting_point()

def p(x,y,p):
    res=1
    x=x%p
    while(y>0):
        if(y&1):
            res=(res*x)%p
        y=y>>1
        x=(x*x)%p
    return res
def nod(x):
    i=0
    while(x):
        x//=10
        i+=1
    return i
def ltd(n):
    temp=1
    for i in range(1,3):
        temp*=10
        temp=p(5,n,temp)
    for i in range(2-nod(temp)):
        print(0,end="")
    if temp:
        print(temp)
def __starting_point():
    n=int(input())
    ltd(n)
__starting_point()

# cook your dish here
n=int(input())
print(25)

b = int(input())
print(int(25))

print(25)


# cook your dish here
n=int(input())
print(25)


n=int(input())
print(25)

# cook your dish here
print('25')

# cook your dish here
n=eval(input())
print(25)


# cook your dish here
num=int(input())
print(25)


# cook your dish here
def power(x, y, p): 
  
    res = 1 # Initialize result 
  
    x = x % p # Update x if it is more  
              # than or equal to p 
  
    while (y > 0): 
  
        # If y is odd, multiply 
        # x with result 
        if (y & 1): 
            res = (res * x) % p 
  
        # y must be even now 
        y = y >> 1 # y = y/2 
        x = (x * x) % p 
          
    return res 
  
# function to calculate 
# number of digits in x 
def numberOfDigits(x): 
  
    i = 0
    while (x): 
        x //= 10
        i += 1
      
    return i 
  
# function to print  
# last 2 digits of 2^n 
def LastTwoDigit(n): 
  
    """print("Last " + str(2) + 
          " digits of " + str(2), end = "") 
    print("^" + str(n) + " = ", end = "")"""
  
    # Generating 10^2 
    temp = 1
    for i in range(1, 3): 
        temp *= 10
  
    # Calling modular exponentiation 
    temp = power(5, n, temp) 
  
    # Printing leftmost zeros.  
    # Since (2^n)%2 can have digits  
    # less then 2. In that case we  
    # need to print zeros 
    for i in range(2 - numberOfDigits(temp)): 
        print(0, end = "") 
  
    # If temp is not zero then print temp 
    # If temp is zero then already printed 
    if temp: 
        print(temp) 
  
# Driver Code 
def __starting_point(): 
    n = int(input())
    LastTwoDigit(n)
    
__starting_point()

a=int(input())
print("25")


n=int(input())

print("25")

n=int(input())

print("25")

# cook your dish here
n=int(input())

print(25)

# cook your dish here
n = int(input())

print(25)