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 fasterthanlight ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define endl "\n"
#define int long long
#define vi vector<int>
#define pb push_back
#define mod 1000000007
#define MAX 1e9
#define MIN -1e9
#define hi "visited\n" //sasta debugging
using namespace std;

//author::@whohet-->Het Patel

void solve()
{
	string s = "";
	int n;
	cin >> n;
	int k;
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		s += ('a' + i);
		for (int j = i + 1; j < k; j++)
		{
			s += ('a' + i);
			s += ('a' + j);
		}
	}
	int m = s.length();
	for (int i = 0; i < n; i++)
		cout << s[i % m];
}

signed main()
{
	fasterthanlight;
	int T; T = 1;
	//cin >> T;
	while (T--)
	{
		solve();
	}

	//cout << "\n\n"<<"Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " seconds.\n";
	return 0;
}