The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."

Could you pass the interview in the machine vision company in IT City?


-----Input-----

The only line of the input contains a single integer n (2 ≤ n ≤ 2·10^18) — the power in which you need to raise number 5.


-----Output-----

Output the last two digits of 5^{n} without spaces between them.


-----Examples-----
Input
2

Output
25

Difficulty: interview

Starter code:

Solutions:
n=int(input())
print(pow(5,n,100))


import sys

n = int(sys.stdin.readline())

print(25)

print(25)

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


n = int(input())
print(pow(5, n, 100))
        
        
    


print(25)

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


input()
print(25)


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


print("25")

print ("25")

print(pow(5, int(input()), 100))

print(25, end='')

print(25)

print(25)

print(25)

n = input()
print("25")

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

n=int(input())
print(pow(5,n,100))




n = int(input())
if n == 1:
    print(5)
else:
    print(25)

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


import re, sys, string, operator, functools, fractions, collections
sys.setrecursionlimit(10**7)
dX= [-1, 1, 0, 0,-1, 1,-1, 1]
dY= [ 0, 0,-1, 1, 1,-1,-1, 1]
RI=lambda x=' ': list(map(int,input().split(x)))
RS=lambda x=' ': input().rstrip().split(x)
mod=int(1e9+7)
eps=1e-6
#################################################
print(25)







n = int(input())
print(5**n if n < 2 else 25)

print(pow(5, int(input()), 100))


3

n = int(input())

print("25")