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 ≤ 2·1018) — the power in which you need to raise number 5.

Output

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

Examples

Input

2


Output

25
'''
t= input()
lng= len(t)
li1=[]; li2=[]
for i in range(lng):
    if t[i]!='a':
        li1.append(t[i])
    elif t[i]=='a':
        li2.append(i)
        
aa= ''.join(li1)
if len(aa)==0:
    print(t); exit(0)
if len(aa)%2==1:
    print(':('); exit(0)
if len(aa)%2==0:
    #print(123)
    l= int(len(aa)/2); lp= l#; print(l)
    ll= aa[l:]; lll=aa[0:l] #; print(ll)
    if ll!=lll:
        print(':('); exit(0)
    if ll not in t:
        print(':('); exit(0)
    tp= t[::-1]; tc= ll[::-1]#; print(tp,tc)
    if tp.find(tc)!=0:
        print(':('); exit(0)
    if tp.find(tc)==0:
        ul= len(tc)
        lu= tp[ul:][::-1]
        print(lu); exit(0)
        
   '''
#print(ord("T")-ord('A')+1)
'''
def fk (a):
    if a=='ACTG':
        return 0
   # print(a)
    cv=0
   # cv+= min(abs(ord(a[0])-ord('A')), (ord('A')-ord('A')+1+ 26- ord(a[0])))
    cv+= (min(abs(ord(a[0])-ord('A')), (ord('A')-ord('A')+1+ 26- (ord(a[0])-ord('A')+1) ), (ord(a[0])-ord('A')+ abs(27- (ord("A")-ord('A')+1)) ) ))
    
   # cv+= min(abs(ord(a[1])-ord('C')), (ord('C')-ord('A')+1+ 26- ord(a[1])))
    cv+= (min(abs(ord(a[1])-ord('C')), (ord('C')-ord('A')+1+ 26- (ord(a[1])-ord('A')+1) ), (ord(a[1])-ord('A')+ abs(27- (ord("C")-ord('A')+1)) ) ))
    
    #cv+= min(abs(ord(a[2])-ord('T')), (ord('T')-ord('A')+1+ 26- ord(a[2])))#; print(abs(ord(a[2])-ord('T')), (ord('T')-ord('A')+1+ 26- ord(a[2])))
    cv+= (min(abs(ord(a[2])-ord('T')), (ord('T')-ord('A')+1+ 26- (ord(a[2])-ord('A')+1) ), (ord(a[2])-ord('A')+ abs(27- (ord("T")-ord('A')+1)) ) ))
    
    #cv+= min(abs(ord(a[3])-ord('G')), (ord('G')-ord('A')+1+ 26- ord(a[3])))
    cv+= (min(abs(ord(a[3])-ord('G')), (ord('G')-ord('A')+1+ 26- (ord(a[3])-ord('A')+1) ), (ord(a[3])-ord('A')+ abs(27- (ord("G")-ord('A')+1)) ) ))
    
    return cv

n= int(input()); s=input(); li=[]
for i in range(n-4+1):
    ab= s[i:i+4]
    li.append(fk(ab))
li.sort()
print(li[0])
'''
'''
n= int(input())
li= list(map(str, input().split())); li= li*2; li= ''.join(li); li= li.replace('0',' ')
li1= li.split()
try:
    print(len(max(li1)))
except:
    print(0)'''

'''
from math import *

def prmfct(n):
    li=[]
    while n%2==0:
        li.append(2)
        n= n/2
    for i in range(3,int(sqrt(n))+1,2):
        while n%i==0:
            li.append(int(i))
            n=int(n/i)
    if n>2:
        li.append(int(n))
        
        
    return li


n,m= input().split(); n,m= int(n), int(m)

if m%n!=0:
    print(-1)
else:
    lim= prmfct(int(m/n))
    li= list(set(lim))
    if len(li)==2 and li[0]==2 and li[1]==3:
        print(len(lim))
    elif len(li)==1 and (li[0]==2 or li[0]==3):
        print(len(lim))
    elif len(li)==0:
        print(0)
    else:
        print(-1)'''
'''
from math import *
a,b= input().split(); a,b= int(a), int(b)
print(factorial(min(a,b)))'''

n=int(input()); li=[6,8,4,2]
print(25)
        




    















'''
Abdul Hamid Shaodagor High School   Joni Kanti Pal  01306261062 Redwan Haque Ratul  01846509454
Hasne Hena Girls High School   Saiyeda Shahinur Parveen    01819858004 Habibur Rahman  01825374474
       ''' 














        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        