#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<map>
#define int long long 
using namespace std;
const int N = 3e5+10, mo=1e9+7;
int n, k;
char s[N];
signed main()
{
	scanf("%lld%lld", &n, &k);
	s[1]='a';
	int p=0, cnt=0;
	if(k==1)
	{
		for(int i=1; i<=n; ++i)
		{
			printf("a");
		}
		return 0;
	}
	while(1)
	{
		if(cnt==k-1)
		{
			s[++p]=97+k-1;
			s[++p]='a';
			break;
		}
		s[++p]=cnt+97;
		for(int i=1; i<=2*(k-1-cnt); ++i)
		{
			if(i%2) s[++p]=cnt+97;
			else s[++p]=cnt+i/2+97;
		}
		if(p>=n) break;	
		cnt++;
	}
	int tot=0;
	printf("%c", s[1]); tot+=1;
	while(1)
	{
		for(int i=2; i<=p; ++i) 
		{
			tot++;
			if(tot>n) return 0;
			printf("%c", s[i]);
		}
	}
}
### Prompt

Generate a Cpp solution to the following problem:

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

### Response

```cpp
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<map>
#define int long long 
using namespace std;
const int N = 3e5+10, mo=1e9+7;
int n, k;
char s[N];
signed main()
{
	scanf("%lld%lld", &n, &k);
	s[1]='a';
	int p=0, cnt=0;
	if(k==1)
	{
		for(int i=1; i<=n; ++i)
		{
			printf("a");
		}
		return 0;
	}
	while(1)
	{
		if(cnt==k-1)
		{
			s[++p]=97+k-1;
			s[++p]='a';
			break;
		}
		s[++p]=cnt+97;
		for(int i=1; i<=2*(k-1-cnt); ++i)
		{
			if(i%2) s[++p]=cnt+97;
			else s[++p]=cnt+i/2+97;
		}
		if(p>=n) break;	
		cnt++;
	}
	int tot=0;
	printf("%c", s[1]); tot+=1;
	while(1)
	{
		for(int i=2; i<=p; ++i) 
		{
			tot++;
			if(tot>n) return 0;
			printf("%c", s[i]);
		}
	}
}
```