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:
#include <bits/stdc++.h>
using namespace std;
#define int                long long int
#define fi                first
#define se                second
#define pb                push_back
#define rep(i,a,n)        for(int i = a;i < n;i++)
#define vec               vector<int>
#define read(v,n)         rep(i,0,n) {cin>>v[i];}
#define pii               pair<int,int>
#define mpp               make_pair(a,b)
#define write(a,n)        rep(i,0,n) {cout<<a[i]<<" ";}cout<<endl;
#define all(v)            v.begin(),v.end()
#define srt(v)            sort(v.rbegin(),v.rend())
#define scn(a)            cin>>a
#define nl                endl
#define dbug(x)          cout << (#x) << " is " << (x) << endl
#define gcd(a,b)          __gcd(a,b)
#define bitcount(n)       __builtin_popcount(n)
#define IOS               ios_base::sync_with_stdio(false); cin.tie(NULL);
#define mod               1000000007

void solve()
{
	int n;
	cin >> n;
	int k;
	cin >> k;
	string ans = "";
	for (int i = 0; i < k; i++)
	{

		ans += (i + 'a');
		for (int j = i + 1; j < k; j++)
		{
			ans += i + 'a';
			ans += j + 'a';
		}
	}
	string s = ans;

	while (s.size() < n)
	{
		s += ans;
	}
	for (int i = 0; i < n; i++)
	{
		cout << s[i];
	}
	cout << nl;
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	IOS
	int t = 1;
	//cin >> t;
	while (t-- > 0)
	{
		solve();
	}
#ifndef ONLINE_JUDGE
	cout << "\nTime Elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " sec\n";
#endif
}