Let's define the cost of a string s as the number of index pairs i and j (1 ≤ i < j < |s|) such that s_i = s_j and s_{i+1} = s_{j+1}.

You are given two positive integers n and k. Among all strings with length n that contain only the first k characters of the Latin alphabet, find a string with minimum possible cost. If there are multiple such strings with minimum cost — find any of them.

Input

The only line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 26).

Output

Print the string s such that it consists of n characters, each its character is one of the k first Latin letters, and it has the minimum possible cost among all these strings. If there are multiple such strings — print any of them.

Examples

Input


9 4


Output


aabacadbb


Input


5 1


Output


aaaaa

Input


10 26


Output


codeforces
n,k=[int(x) for x in input().split(' ')]

if(k==1):
	for i in range(n):
		print("a",end="")
	print()
else:

	a=[]
	for i in range(k):
		a.append([ tuple([i,j]) for j in range(k) ])

	stat=[ [(-1) for j in range(k)] for i in range(k) ]

	basicString=[]

	def dfs(currI,currJ):
		global stat
		# display()
		if(stat[currI][currJ]==k*k):
			# path is found
			basicString.append([currI,currJ])
			return(1)
		endingnode=currJ
		for itr in range(k):
			if(stat[endingnode][itr]==-1):
				stat[endingnode][itr]=(stat[currI][currJ]+1)
				res=dfs(endingnode,itr)
				if(res==1):
					basicString.append([currI,currJ])
					return(1)
				stat[endingnode][itr]=-1
		return(0)

	stat[0][1]=1;
	dfs(0,1)
	basicString.reverse()

	# print(basicString)

	bsT=[]
	for x in basicString:
		bsT.append(x[0])

	# print(bsT)

	for i in range(n):
		print(  chr((bsT[i%len(bsT)]) + 97)  ,end="")
	print()
