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>
using namespace std;
#define ll long long
#define ld long double
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vlli vector<ll>
#define vi vector<int>
#define vs vector<string>
#define vplli vector< pair< ll,ll> >
#define plli pair< ll,ll >
#define vps vector< pair< string, string> >
#define vpi vector< pair< int, int> >
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define forn(i,a,n) for(ll i=a;i<n;i++)
#define forr(i,n,a) for(ll i=n-1;i>=a;i--)
#define scan(arr,a,n) for(ll i=(a);i<(n);i++)cin>>(arr)[i];
#define print(arr,a,n) for(ll i=(a);i<(n);i++)cout<<(arr)[i]<<" ";

const ll inf = 1e18;

ll add(ll x, ll y) {ll res = x + y; return (res >= mod ? res - mod : res);}
ll mul(ll x, ll y) {ll res = x * y; return (res >= mod ? res % mod : res);}
ll sub(ll x, ll y) {ll res = x - y; return (res < 0 ? res + mod : res);}
ll power(ll x, ll y) {ll res = 1; x %= mod; while (y) {if (y & 1)res = mul(res, x); y >>= 1; x = mul(x, x);} return res;}
ll mod_inv(ll x) {return power(x, mod - 2);}


int main(){
    fast;
    #ifndef ONLINE_JUDGE
    freopen("input1.txt","r",stdin);
    freopen("output1.txt","w",stdout);
    #endif
    ll n, k;
    cin>>n>>k;
    if(k==1 or n==1){
        forn(i, 0, n){
            cout<<'a';
        }
        cout<<endl;
        return 0;
    }
    string ans = "";
    forn(i, 0, k){
        char c = (char)('a'+i);
        for(char cc = c; cc<(char)('a'+k);cc++){
            ans+=cc;
            if(cc != (char)('a'+k-1)){
                ans+=c;
            }
        }
    }
    ll x = ans.size();
    ll z = n/x;
    string s = "";
    forn(i, 0, z){
        s+=ans;
    }
    forn(i, 0, n%x){
        s+=ans[i];
    }
    cout<<s;
}