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

Solution (in CPP):

Solution:
// Problem: D. Строка минимальной стоимости
// Contest: Codeforces - Educational Codeforces Round 107 (рейтинговый для Див. 2)
// URL: https://codeforces.com/contest/1511/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define ll long long 
#define mp make_pair
#define pb push_back
#define ve vector
#define fi first
#define se second
#define sz(s) (int)s.size()
#define nl "\n"
//
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define rop(i, n) for(int i = 0; i < n; i++)
#define r1p(i, n) for(int i = 1; i <= n; i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rev(x) reverse(all(x))
#define srt(x) sort(all(x))
//
#define yep cout << "YES" << nl
#define nope cout << "NO" << nl
#define retr return
//
using namespace std;
typedef ve<int> vi;
typedef ve<ll> vll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
//
const int MOD = 1e9;
const int mxN = 1e5 + 5;
void solve() {
	
}
void cf() {int tt;cin >> tt;while(tt--) {solve();}}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//

	//freopen("cin.in", "r", stdin);
	//freopen("cin.out", "w", stdout);
	int n, k;
	cin >> n >> k;
	string ans = "";
	for(int i = 0; i < k; i++) {
		ans += char(int('a') + i);
		for(int j = i + 1; j < k; j++) {
			ans += char(int('a') + i);
			ans += char(int('a') + j);
		}
	}
	while(sz(ans) < n)
		ans += ans;
	ans.resize(n);
	cout << ans;
	
	
}