#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
using namespace std;

typedef long long ll;
const int mod = 1000000007;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int n, k; cin >> n >> k; 
	vector<char> V(k*2, -1);
	for (int i = 0; i < k; i++) { V[i] = 'a' + i; }
	for (int i = k; i < 2 * k; i++) { V[i] = V[i - k]; }
	string s; s = "";
	for (int i = 0; i < k; i++) {
		s += V[i];
		for (int j = i+1; j < k; j++) {
			s += V[i]; s += V[j];
		}
	}

	while (s.size() < n) { s += s; }
	cout << s.substr(0, n) << '\n';
	cout << '\n';
	return 0;
}

### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
using namespace std;

typedef long long ll;
const int mod = 1000000007;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int n, k; cin >> n >> k; 
	vector<char> V(k*2, -1);
	for (int i = 0; i < k; i++) { V[i] = 'a' + i; }
	for (int i = k; i < 2 * k; i++) { V[i] = V[i - k]; }
	string s; s = "";
	for (int i = 0; i < k; i++) {
		s += V[i];
		for (int j = i+1; j < k; j++) {
			s += V[i]; s += V[j];
		}
	}

	while (s.size() < n) { s += s; }
	cout << s.substr(0, n) << '\n';
	cout << '\n';
	return 0;
}

```