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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
public class Questions{
	static long mod = 1000000007;
	
	static long power(long x, long y) {
		long res = 1;
		
		while (y > 0) {
			if ((y & 1) != 0)
				res = (res * x) % mod;
			y = y >> 1;
			x = (x * x) % mod;
		}
		return res;
	}
	
	static int gcd(int a, int b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}
	
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
		
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		int t = sc.nextInt();
		while(t-- != 0) {
			int len = sc.nextInt();
			if(len <= 4) {
				if(len == 1) {
					System.out.println("a");
				}else if(len == 2) {
					System.out.println("ab");
				}else if(len == 3) {
					System.out.println("abc");
				}else {
					System.out.println("abcd");
				}
			}else {
				if(len % 2 == 1) {
					int afterMiddle = len-2;
					int left = afterMiddle / 2;
					int right = left+1;
					StringBuilder str = new StringBuilder();
					while(left != 0) {
						str.append('a');
						left--;
					}
					str.append("bc");
					while(right != 0) {
						str.append('a');
						right--;
					}
					System.out.println(str.toString());
				}else {
					int afterMiddle = len-1;
					int left = afterMiddle / 2;
					int right = left+1;
					StringBuilder str = new StringBuilder();
					while(left != 0) {
						str.append('a');
						left--;
					}
					str.append("b");
					while(right != 0) {
						str.append('a');
						right--;
					}
					System.out.println(str.toString());
				}
			}
		}
	}
}