#include<bits/stdc++.h>
using namespace std;
#define int 		long long
#define pb 			push_back
#define all(a) 		a.begin(),a.end()
#define pii 		pair<int,int>
#define fi 			first
#define se 			second
#define rep(i,k,n) 	for(int i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
const int M=1e9+7,N=205001;

int mm(int x,int y){x%=M,y%=M;return (x*y)%M;}//Modular Multiply
int po(int x,int y){ if(!y)return 1;int a=po(x,y/2)%M;if(y%2)return mm(a,mm(a,x));return mm(a,a);}//(x**y)%M



void solve() {
	int n,k;
	cin>>n>>k;
	string a,t,p;
	rep(i,0,k) a+=(i+'a');
	rep(i,0,k) rep(j,i,k) {
		t+=a[i];t+=a[j];	
	}
	rep(i,0,t.size()){
		if(i>1 and t[i]==t[i-2] and t[i]==t[i-1]) continue;
		p+=t[i];
	}
	k=p.size();
	rep(i,0,n) cout<<p[i%k];
}


signed main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t=1;
	//cin>>t;
	while(t--) solve();
	return 0;
}





### Prompt

Create a solution in Cpp for 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;
#define int 		long long
#define pb 			push_back
#define all(a) 		a.begin(),a.end()
#define pii 		pair<int,int>
#define fi 			first
#define se 			second
#define rep(i,k,n) 	for(int i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
const int M=1e9+7,N=205001;

int mm(int x,int y){x%=M,y%=M;return (x*y)%M;}//Modular Multiply
int po(int x,int y){ if(!y)return 1;int a=po(x,y/2)%M;if(y%2)return mm(a,mm(a,x));return mm(a,a);}//(x**y)%M



void solve() {
	int n,k;
	cin>>n>>k;
	string a,t,p;
	rep(i,0,k) a+=(i+'a');
	rep(i,0,k) rep(j,i,k) {
		t+=a[i];t+=a[j];	
	}
	rep(i,0,t.size()){
		if(i>1 and t[i]==t[i-2] and t[i]==t[i-1]) continue;
		p+=t[i];
	}
	k=p.size();
	rep(i,0,n) cout<<p[i%k];
}


signed main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t=1;
	//cin>>t;
	while(t--) solve();
	return 0;
}





```