#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int mod = 1e9+7;
const int inf = 1e9+100;
const int N = 3e5+5;


int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

    int n,k;
    cin>>n>>k;
    string cur = "aa";
    for(int i=2; i<=k; i++){
        char ch = char('a'+i-1);
        cur += ch;
        for(int j=i-1; j>=1; j--){
            cur += ch;
            cur += char('a'+j-1);
        }
    }

    string ans;
    for(int i=0, j=0; i<n; i++, j++){
        if(j==cur.size()) j = 1;
        ans += cur[j];
    }

    cout<<ans<<"\n";
}
/*
*/

### Prompt

Develop a solution in CPP to the problem described below:

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;
typedef long long LL;
typedef pair<int,int>pii;
const int mod = 1e9+7;
const int inf = 1e9+100;
const int N = 3e5+5;


int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

    int n,k;
    cin>>n>>k;
    string cur = "aa";
    for(int i=2; i<=k; i++){
        char ch = char('a'+i-1);
        cur += ch;
        for(int j=i-1; j>=1; j--){
            cur += ch;
            cur += char('a'+j-1);
        }
    }

    string ans;
    for(int i=0, j=0; i<n; i++, j++){
        if(j==cur.size()) j = 1;
        ans += cur[j];
    }

    cout<<ans<<"\n";
}
/*
*/

```