#include<bits/stdc++.h>
using namespace std;
int a[200005];
int main(){
	int n, k; cin >> n >> k;
	if(k*k >= n) {
		int ans = 0;
		for(int i=1;i<=k;i++) {
			a[++ans] = i;
			for(int j=i+1;j<=k;j++) {
				a[++ans] = i;
				a[++ans] = j;
				if(ans>=n) break;
			}
		}
	}
	else {
		int ans = 0;
		while(ans<=n){
			for(int i=1;i<=k;i++) {
				a[++ans] = i;
				for(int j=i+1;j<=k;j++) {
					a[++ans] = i;
					a[++ans] = j;
					if(ans>=n)break;
				}
				if(ans>=n)break;
			}
			if(ans>=n)break;
		}
	}
	for(int i=1;i<=n;i++) {
		printf("%c",a[i]+'a'-1);
	}
	return 0;
}
### Prompt

Please create a solution in cpp 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<bits/stdc++.h>
using namespace std;
int a[200005];
int main(){
	int n, k; cin >> n >> k;
	if(k*k >= n) {
		int ans = 0;
		for(int i=1;i<=k;i++) {
			a[++ans] = i;
			for(int j=i+1;j<=k;j++) {
				a[++ans] = i;
				a[++ans] = j;
				if(ans>=n) break;
			}
		}
	}
	else {
		int ans = 0;
		while(ans<=n){
			for(int i=1;i<=k;i++) {
				a[++ans] = i;
				for(int j=i+1;j<=k;j++) {
					a[++ans] = i;
					a[++ans] = j;
					if(ans>=n)break;
				}
				if(ans>=n)break;
			}
			if(ans>=n)break;
		}
	}
	for(int i=1;i<=n;i++) {
		printf("%c",a[i]+'a'-1);
	}
	return 0;
}
```