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
#include <bits/stdc++.h>
#define N 1000000007
using namespace std;
 unordered_map<long long,long long>mp;
 int upIndex[26]={0};
 int downIndex[26]={0};
 vector<int>up[26];
 vector<int>down[26];
 string s="";
 void h(int index,int k)
 {
     
         if(upIndex[index]<up[index].size()){
             int kk=up[index][upIndex[index]];
             upIndex[index]++;
             s+=char('a'+kk);
             h(kk,k);
         }
         else
         {
             if(downIndex[index]>=down[index].size())
             {
                 return;
             }
             int kk=down[index][downIndex[index]];
             downIndex[index]++;
             s+=char('a'+kk);
             h(kk,k);
         }
     
 }
int main() {
     ios_base::sync_with_stdio(false);
    cin.tie(NULL);
   
    int n,k;
    cin>>n>>k;
    
    // if(k)
    // cout<<'a';
    s='a';
    for(int i=0;i<k;i++){
        for(int j=i;j<k;j++)
        {
            up[i].push_back(j);
        }
        for(int k=i-1;k>=0;k--)
        {
            down[i].push_back(k);
        }

    }
    h(0,k);
    // cout<<s<<endl;
    if(n<=s.size())
    {
        cout<<s.substr(0,n);
    }
    else
    {
        if(k==1)
        {
            for(int i=0;i<n;i++)
            cout<<'a';
            
        }
        else
        {
            cout<<s;
            
            int l=s.size();
            
            s=s.substr(1);
            int mm=s.size();
            l+=mm;
            while(l<=n)
            {
                cout<<s;
                l+=mm;
            }
            l-=mm;
            n-=l;
            cout<<s.substr(0,n);
        }
    }

}