// Arup Guha
// 3/29/2015
// Solution to 2015 NAIPC Problem J: Zig Zag Nametag

import java.util.*;

public class zigzag {

	public static void main(String[] args) {

		Scanner stdin = new Scanner(System.in);
		int n = stdin.nextInt();

		int len = (n+24)/25+1;

		// Chars in answer due to azaza answer...
		char[] res = new char[len];
		res[0] = 'a';

		// Easy case - but different pattern than the rest.
		if (len == 2) {
			res[1] = (char)('a' + n);
		}

		// Typical greedy.
		else {

			// This is key - make this as "small" as possible without adding characters.
			int fromZ = n%25 == 0 ? 0 : (25 - n%25)/2;
			res[1] = (char)('z' - fromZ);
			int cur = res[1] - res[0];

			// Just do a regular zig zag here, until the last character.
			for (int i=2; i<len-1; i++) {
				res[i] = i%2 == 0 ? 'a' : 'z';
				cur = cur + Math.abs(res[i]-res[i-1]);
			}

			// Just because I'm lazy, this is the last character calculation.
			int left = n - cur;
			res[len-1] = (len-1)%2 == 0 ? (char)(res[len-2]-left) : (char)(res[len-2]+left);
		}

		// Ta da!
		System.out.println(new String(res));
	}
}