You are given an integer n. Find any string s of length n consisting only of English lowercase letters such that each non-empty substring of s occurs in s an odd number of times. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print a single line containing the string s. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.

Example

Input


4
3
5
9
19


Output


abc
diane
bbcaabbba
youarethecutestuwuu

Note

In the first test case, each substring of "abc" occurs exactly once.

In the third test case, each substring of "bbcaabbba" occurs an odd number of times. In particular, "b" occurs 5 times, "a" and "bb" occur 3 times each, and each of the remaining substrings occurs exactly once.
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class D{
	
	static int cnt(String s) {
		HashSet<String> hs = new HashSet<String>();
		for (int i = 0; i < s.length(); i++) {
			for (int j= i; j < s.length(); j++) {
				hs.add(s.substring(i,j+1));
			}
		}
		out.println(hs);
		return hs.size();
	}
	
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		//new Thread(null, new (), "peepee", 1<<28).start();
		
		int t= readInt();
		while(t-->0) {
			int n =readInt();
			if (n==1) out.println("a");
			else if (n%2==0) {
				StringBuilder front = new StringBuilder(), end = new StringBuilder();
				n-=3;
				int turns = 0;
				while(n>0) {
					n-=2;
					if (turns%2==0) end.append("aa");
					else front.append("aa");
					turns++;
				}
				out.println(front.toString() + "ab" + end.toString());
			}
			else {
				StringBuilder front = new StringBuilder(), end = new StringBuilder();
				n-=3;
				int turns = 0;
				while(n>0) {
					n-=2;
					if (turns%2==0) end.append("aa");
					else front.append("aa");
					turns++;
				}
				out.println(front.toString() + "abc" + end.toString());
			}
		}
		out.close();
	}
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
}