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
#pragma GCC optimize("Ofast")
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef long double ld;
#define fast ios_base::sync_with_stdio(false); cin.tie(nullptr);
#define all(x) x.begin(), x.end()
#define F first
#define S second
#define pb push_back
#define pl pair<ll, ll>
#define vl vector<ll>
#define vi vector<int>
#define endl '\n'

template <typename T, typename TT>
inline ostream &operator<<(ostream &os, const pair<T, TT> &t) {
	return os << t.first << " " << t.second;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &t) {
	for (auto i : t) os << i << " ";
	return os;
}
template <typename T>
inline istream &operator>>(istream &is, vector<T> &v) {
	for (T &t : v) is >> t;
	return is;
}
template <typename T1, typename T2>
inline istream &operator>>(istream &is, pair<T1, T2> &t) {
	is >> t.first >> t.second;
	return is;
}

#ifdef LOCAL
#define debug(args...) (Debugger()), args
class Debugger {
public:
	bool first;
	string separator;
	Debugger(const string &_separator = ", ") : first(true), separator(_separator) {}
	template <typename ObjectType>
	Debugger &operator, (const ObjectType &v) {
		if (!first) cerr << separator;
		cerr << v;
		first = false;
		return *this;
	}
	~Debugger() { cerr << endl; }
};
#else
#define debug(args...)
#endif

const ll mod = 1e9 + 7;

void solve() {
	ll n, k;
	cin >> n >> k;
	if (n == 1) {
		cout << 'a' << endl;
		return;
	}
	string ans;
	for (int i = 0; i < k - 1; i++) {
		ans += char(i + 'a');
		for (int j = i + 1; j < k; j++)
			ans += char(i + 'a'), ans += char(j + 'a');
	}
	ans += char(k - 1 + 'a');
	string aa;
	while (aa.size() < n) aa += ans;
	cout << aa.substr(0, n) << endl;
}

int main()
{
	fast;
	int T = 1;
	// cin >> T;
	while (T--) {
		solve();
	}
	return 0;
}
