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
//Implemented By Aman Kotiyal Date:-12-Apr-2021  Time:-7:55:19 pm
import java.io.*;
import java.util.*;

public class ques5
{
	public static void main(String[] args)throws Exception{ new ques5().run();} 
	long mod=1000000000+7;
	
	void solve() throws Exception
	{
			int n=ni();
			int k=ni();
			
			char lim=(char)('a'+(k-1));
			char s[]=new char[n];
			s[0]='a';
			int curr=1;
			
			String miss=lim+"a";
			HashSet<String> set=new HashSet<>();
			char upadd='a';
			char add='a';
			while(curr<n)
			{
				String wd=s[curr-1]+""+add;
				if(wd.equals(miss))
					add=(char)(add+1);
				if(add>lim)
				{
					add='a';
					upadd=(char)(upadd+1);
				}
				if(upadd>lim)
				{
					break;
				}
				
				wd=s[curr-1]+""+add;
				
				if(!set.contains(wd))
				{
					s[curr]=add;
					curr++;
					set.add(wd);
					add='a';
				}
				else
				{
					add=(char)(add+1);
					if(add>lim)
					{
						add='a';
						upadd=(char)(upadd+1);
					}
					if(upadd>lim)
					{
						break;
					}
				}
			}
			if(curr<n)
			{
				s[curr]='a';
				curr++;
			}
			if(curr==n)
			{
				for (int i = 0; i < s.length; i++) {
					out.print(s[i]);
				}
				out.println();
			}
			else
			{
				for (int i = 0; i <curr; i++) {
					out.print(s[i]);
				}
				int in=curr+1;
				int pp=1;
				while(in<=n)
				{
					out.print(s[pp]);
					pp++;
					if(pp==curr)
						pp=1;
					in++;
				}
				out.println();
			}
	}
	
	
	/*FAST INPUT OUTPUT & METHODS BELOW*/
	
	private byte[] buf=new byte[1024];
	private int index;
	private InputStream in;
	private int total;
	private SpaceCharFilter filter;
	PrintWriter out;
	
	int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
	long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
	int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
	long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
	void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
	void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
	String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
	
	void shuffle(int a[]) {
		ArrayList<Integer> al = new ArrayList<>();
		for(int i=0;i<a.length;i++) 
			al.add(a[i]);
		
		Collections.sort(al);
		for(int i=0;i<a.length;i++) 
			a[i]=al.get(i);
	}
	long lcm(long a,long b)
	{
		return (a*b)/(gcd(a,b));
	}
	
	int gcd(int a, int b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b%a, a); 
	} 
	long gcd(long a, long b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b%a, a); 
	}
	/* for (1/a)%mod = ( a^(mod-2) )%mod  ----> use expo to calc -->(a^(mod-2)) */
	long expo(long p,long q)  /*  (p^q)%mod   */
	{
		long z = 1;
		while (q>0) {
			if (q%2 == 1) {
				z = (z * p)%mod;
			}
			p = (p*p)%mod;
			q >>= 1;
		}
		return z;
	}
	void run()throws Exception
	{
		in=System.in; out = new PrintWriter(System.out);
		solve();
		out.flush();
	}
	private int scan()throws IOException
	{
		if(total<0)
			throw new InputMismatchException();
		if(index>=total)
		{
			index=0;
			total=in.read(buf);
			if(total<=0)
				return -1;
		}
		return buf[index++];
	}
	private int ni() throws IOException 
	{
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = scan();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = scan();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
	private long nl() throws IOException 
	{
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = scan();
		}
		
		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = scan();
		}
	}
	private double nd() throws IOException{
		return Double.parseDouble(ns());
	}
	private String ns() throws IOException {
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		StringBuilder res = new StringBuilder();
		do {
			if (Character.isValidCodePoint(c))
				res.appendCodePoint(c);
			c = scan();
		} while (!isSpaceChar(c));
		return res.toString();
	}
	private String nss() throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		return br.readLine();
	}
	private char nc() throws IOException 
	{
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		return (char) c;
	}
	private boolean isWhiteSpace(int n)
	{
		if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
			return true;
		return false;
	}
	private boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return isWhiteSpace(c);
	}
	private interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}