The cows have just learned what a primitive root is! Given a prime *p*, a primitive root <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f203bd9f1553a2d666452cec7d6f1f4f02a817fb.png" style="max-width: 100.0%;max-height: 100.0%;"/> is an integer *x* (1 ≤ *x* &lt; *p*) such that none of integers *x* - 1, *x*<sup class="upper-index">2</sup> - 1, ..., *x*<sup class="upper-index">*p* - 2</sup> - 1 are divisible by *p*, but *x*<sup class="upper-index">*p* - 1</sup> - 1 is. 

Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime *p*, help the cows find the number of primitive roots <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f203bd9f1553a2d666452cec7d6f1f4f02a817fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The input contains a single line containing an integer *p* (2 ≤ *p* &lt; 2000). It is guaranteed that *p* is a prime.
p=int(input())
global res
res=0
global prim_root
prim_root=[]
for x in range(1,p):
	for i in range(1,p):
		if(pow(x,i)-1!=0):
			if ((pow(x,i)-1)%p==0):
					res+=1
		print ('x=',x,'i=',i,'pow=',((pow(x,i)-1)%p==0),'ic=',(i==p-1),'res=',res)
		if (res==1):
			if (i==p-1):
				prim_root.append(x)
				res=0
	res=0
print (list(prim_root))
			
	# if (res==1):
	# 	prim_root.append(x)
# print (list(prim_root))