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For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:

  * Remove any two elements s and t from the set, and add their concatenation s + t to the set. 



The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.

Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.

Input

The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.

Output

Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.

Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.

Examples

Input

12


Output

abababab


Input

3


Output

codeforces

Note

For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:

  * {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; 
  * {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; 
  * {"abab", "a", "b", "a", "b"}, with a cost of 1; 
  * {"abab", "ab", "a", "b"}, with a cost of 0; 
  * {"abab", "aba", "b"}, with a cost of 1; 
  * {"abab", "abab"}, with a cost of 1; 
  * {"abababab"}, with a cost of 8. 



The total cost is 12, and it can be proved to be the minimum cost of the process.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long n = in.nextLong();

            long[] t = new long[2000];
            for (int i = 0; i < 2000; i++) {
                t[i] = (long) i * (i - 1) / 2;
            }

            StringBuilder sb = new StringBuilder();
            int idx = 0;
            for (int i = 1999; i >= 3; i--) {
                if (n >= t[i]) {
                    for (int j = 0; j < i; j++) {
                        sb.append((char) ('a' + idx));
                    }
                    idx++;
                    n -= t[i];
                }
            }

            for (int i = 0; i < n; i++) {
                sb.append((char) ('a' + idx));
                sb.append((char) ('a' + idx));
                idx++;
            }

            out.println("".equals(sb.toString()) ? "a" : sb.toString());
        }

    }

    static class InputReader {
        BufferedReader in;
        StringTokenizer tok;

        public String nextString() {
            while (!tok.hasMoreTokens()) {
                try {
                    tok = new StringTokenizer(in.readLine(), " ");
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
            }
            return tok.nextToken();
        }

        public long nextLong() {
            return Long.parseLong(nextString());
        }

        public InputReader(InputStream inputStream) {
            in = new BufferedReader(new InputStreamReader(inputStream));
            tok = new StringTokenizer("");
        }

    }
}

