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.util.*;
import java.io.*;

public class D_1554 {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int t = sc.nextInt();
		while(t-->0) {
			int n = sc.nextInt();
			if(n == 1)
				pw.println("a");
			else if(n == 2)
				pw.println("ab");
			else if(n == 3)
				pw.println("abc");
			else {
				String mid = "b";
				if(n % 2 == 1)
					mid = "bc";
				n -= n % 2 == 0 ? 1 : 2;
				StringBuilder sb = new StringBuilder("");
				int q = n / 2;
				for(int i = 0; i < q; i++)
					sb.append("a");
				sb.append(mid);
				for(int i = 0; i < n - q; i++)
					sb.append("a");
				
				pw.println(sb);
			}
		}
		
		pw.flush();
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] array = new int[n];
			for (int i = 0; i < n; i++)
				array[i] = nextInt();
			return array;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] array = new Integer[n];
			for (int i = 0; i < n; i++)
				array[i] = new Integer(nextInt());
			return array;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++)
				array[i] = nextLong();
			return array;
		}

		public double[] nextDoubleArray(int n) throws IOException {
			double[] array = new double[n];
			for (int i = 0; i < n; i++)
				array[i] = nextDouble();
			return array;
		}

		public static int[] shuffle(int[] a) {
			int n = a.length;
			Random rand = new Random();
			for (int i = 0; i < n; i++) {
				int tmpIdx = rand.nextInt(n);
				int tmp = a[i];
				a[i] = a[tmpIdx];
				a[tmpIdx] = tmp;
			}
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}
