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
/**
*!! Abhimanyu Bano !!
*!!कर्म करो फल की चिंता मत करो !!
* @Author  : Abhinandan Mishra
*
**/
#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define pb        push_back
#define ppb       pop_back
#define pf        push_front
#define ppf       pop_front
#define all(x)    (x).begin(),(x).end()
#define uniq(v)   (v).erase(unique(all(v)),(v).end())
#define f        first
#define s        second
#define pll       pair<ll,ll>
#define vll        vector<ll>
#define vpll        vector<pair<ll,ll>>
#define mll          map<ll,ll>
#define umll          unordered_map<ll,ll>
#define usll          unordered_set<ll>
#define debug(x) cerr<<'\n'<<(#x)<<" is "<<(x)<<endl;
#define sll          set<ll>
#define FILL(arr)         for(ll i=0;i<n;i++)cin>>arr[i];
#define fill(arr)         for(ll i=0;i<n;i++){ ll k; cin>>k;arr.pb(k); }
#define Print(arr)         for(ll i=0;i<n;i++)cout<<arr[i]<<' ';
#define charcountmap(s)         for(ll i=0;i<s.length();i++)mp[s[i]]++;
#define rep(i,n)   for(ll i=0;i<n;i++)
#define repe(i,a,b)   for(ll i=a;i<=b;i++)
#define mem1(a)      memset(a,-1,sizeof(a))
#define mem0(a)      memset(a,0,sizeof(a))
#define INF 100000000000000000
#define mod 1000000007
#define esp 10e-7
const int mx=1e5+7;


ll gcd(ll a, ll b){
if (b == 0)
return a;
return gcd(b, a % b);
}
ll min_used[26];
void solve(){
    rep(i,26){
        min_used[i]=0;
    }
    ll n,k;
    cin>>n>>k;
    ll j=0;
    for(ll i=0;i<n;i++){
            cout<<char('a'+j);
            j=(j+min_used[j]++)%k;
    
    }
    
    cout<<endl;

}

int main(){
    ll t=1;
    while(t--){
        solve();
    }

    return 0;
}