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.
//This code is written by प्रविण शंखपाळ 

//package wizard;

import java.io.*;
import java.math.*;
import java.util.*;
import java.text.*;

public class Luna_Lovegood {

	public static void main(String[] args) {

//		try {

		FastReader fr = new FastReader();
		PrintWriter pt = new PrintWriter(System.out);

		int t = fr.nextInt();

		while (t > 0) {

			int n = fr.nextInt();

			if (n == 1) {
				pt.print("a");
			} else if (n % 2 == 0) {

				for (int i = 0; i < n / 2; i++) {
					pt.print("a");
				}
				pt.print("b");
				for (int i = 0; i < n / 2 - 1; i++) {
					pt.print("a");
				}
			} else {

				for (int i = 0; i < n / 2; i++) {
					pt.print("a");
				}
				pt.print("bc");
				for (int i = 0; i < n / 2 - 1; i++) {
					pt.print("a");
				}

			}
			pt.println();

			t--;

		}

		pt.close();

//		} catch (
//
//		Exception e) {
//			return;
//		}
	}

	static int lower_bound(long[] arr, int begin, int end, long tar) {
		while (begin < end) {
			int mid = begin + (end - begin) / 2;
			if (arr[mid] < tar)
				begin = mid + 1;
			else if (arr[mid] >= tar)
				end = mid;
		}
		return begin;
//		lower bound gives largest index of a number smaller than (if target is not present) or equal to target(if target is present) 
//      for complete array use begin = 0 and end = n
	}

	static int upper_bound(long[] arr, int begin, int end, long tar) {
		while (begin < end) {
			int mid = begin + (end - begin) / 2;
			if (arr[mid] <= tar)
				begin = mid + 1;
			else if (arr[mid] > tar)
				end = mid;
		}

		return begin;
//		upper bound gives smallest index of a number strictly greater than target
//	    for complete array use begin = 0 and end = n
	}

	public static void mergeSort(String[] nameGo) {
		if (nameGo.length > 1) {
			String[] leftGo = new String[nameGo.length / 2];
			String[] rightGo = new String[nameGo.length - nameGo.length / 2];
			for (int so = 0; so < leftGo.length; so++) {
				leftGo[so] = nameGo[so];
			}
			for (int ki = 0; ki < rightGo.length; ki++) {
				rightGo[ki] = nameGo[ki + nameGo.length / 2];
			}
			mergeSort(leftGo);
			mergeSort(rightGo);
			merges(nameGo, leftGo, rightGo);
		}
	}

	public static void merges(String[] nameH, String[] leftH, String[] rightH) {
		int as = 0;
		int bs = 0;
		for (int i = 0; i < nameH.length; i++) {
			if (bs >= rightH.length || (as < leftH.length && leftH[as].compareToIgnoreCase(rightH[bs]) < 0)) {
				nameH[i] = leftH[as];
				as++;
			} else {
				nameH[i] = rightH[bs];
				bs++;
			}
		}
	}

	static void merge(long arr[], int l, int m, int r) {

		int n1 = m - l + 1;
		int n2 = r - m;

		long L[] = new long[n1];
		long R[] = new long[n2];

		for (int i = 0; i < n1; ++i)
			L[i] = arr[l + i];
		for (int j = 0; j < n2; ++j)
			R[j] = arr[m + 1 + j];

		int i = 0, j = 0;

		int k = l;
		while (i < n1 && j < n2) {
			if (L[i] <= R[j]) {
				arr[k] = L[i];
				i++;
			} else {
				arr[k] = R[j];
				j++;
			}
			k++;
		}

		while (i < n1) {
			arr[k] = L[i];
			i++;
			k++;
		}

		while (j < n2) {
			arr[k] = R[j];
			j++;
			k++;
		}
	}

	static void sort(long arr[], int l, int r) {
		if (l < r) {

			int m = l + (r - l) / 2;

			sort(arr, l, m);
			sort(arr, m + 1, r);

			merge(arr, l, m, r);
		}
	}

	static class Pair implements Comparable<Pair> {
		int a, b;

		Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			if (this.a != o.a)
				return Integer.compare(this.a, o.a);
			else
				return Integer.compare(this.b, o.b);
			// return 0;
		}

		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair) o;
				return p.a == a && p.b == b;
			}
			return false;
		}

	}

	static int binarySearch(int arr[], int first, int last, int key) {
		int mid = (first + last) / 2;
		while (first <= last) {
			if (arr[mid] < key) {
				first = mid + 1;
			} else if (arr[mid] == key) {
				return mid;
			} else {
				last = mid - 1;
			}
			mid = (first + last) / 2;
		}
		return -1;
	}

	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;
		}
	}
}
