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.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;

public class Diane {

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			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) {

		FastScanner f = new FastScanner();
		PrintWriter p = new PrintWriter(System.out);
		int t = f.nextInt();
		while (t-- > 0) {
			int n = f.nextInt();
			StringBuffer s = new StringBuffer();
			if (n == 1) {
				p.println("a");
			} else {
				if (n % 2 != 0) {
					for (int i = 0; i < n / 2; i++) {
						s.append('a');
					}
					s.append("bc");
					for (int i = 0; i < n / 2 - 1; i++) {
						s.append('a');
					}
					String res = s.toString();
					p.println(res);
				} else {
					for (int i = 0; i < n / 2; i++) {
						s.append('a');
					}
					s.append("b");
					for (int i = 0; i < n / 2 - 1; i++) {
						s.append('a');
					}
					String res = s.toString();
					p.println(res);
				}
			}
		}
		p.close();

	}

	// will work till array size of 65537---fastest sorting time
//	static long[] radixSort(long[] f) {
//		return radixSort(f, f.length);
//	}
//
//	static long[] radixSort(long[] f, int n) {
//		long[] to = new long[n];
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] >>> 16 & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] >>> 16 & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] >>> 32 & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] >>> 32 & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] >>> 48 & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] >>> 48 & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		return f;
//	}

	// Function to sort an array using quick sort algorithm.
	static void quickSort(long arr[], int low, int high) {
		if (low < high) {
			int p = partition(arr, low, high);
			quickSort(arr, low, p - 1);
			quickSort(arr, p + 1, high);
		}
	}

	static int partition(long arr[], int low, int high) {
		long pivot = arr[high];
		int i = low - 1;
		for (int j = low; j < high; j++) {
			if (arr[j] < pivot) {
				i++;
				long temp = arr[i];
				arr[i] = arr[j];
				arr[j] = temp;
			}
		}
		long temp = arr[i + 1];
		arr[i + 1] = arr[high];
		arr[high] = temp;

		return i + 1;
	}

	/// Ceils
	static long ceil(long x, long m) {
		long res = x / m;
		if (x % m != 0) {
			res++;
		}
		return res;
	}

	// ------------------------------------------------------------------------------------------------
	// makes the prefix sum array
	static long[] prefixSum(long arr[], int n) {
		long psum[] = new long[n];
		psum[0] = arr[0];
		for (int i = 1; i < n; i++) {
			psum[i] = psum[i - 1] + arr[i];
		}
		return psum;
	}
	// ------------------------------------------------------------------------------------------------

	// makes the suffix sum array
	static long[] suffixSum(long arr[], int n) {
		long ssum[] = new long[n];
		ssum[n - 1] = arr[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			ssum[i] = ssum[i + 1] + arr[i];
		}
		return ssum;
	}

//------------------------------------------------------------------------------------------
	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
	// OVERHEADS
	static long m = (long) (1e9 + 7);

	static long binPower(long a, long n, long m) {
		if (n == 0)
			return 1;
		long res = 1;
		while (n > 0) {
			if ((n & 1) != 0) {
				res *= a;
			}
			a *= a;
			n >>= 1;
		}
		return res;
	}

//-------------------------------------------------------------------------------------------

	// gcd
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

//------------------------------------------------------------------------------------------

	// lcm
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

//------------------------------------------------------------------------------------------
	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
	// O(LOGn)
	static int setBits(int n) {
		int count = 0;
		while (n > 0) {
			n = n & (n - 1);
			count++;
		}
		return count;
	}
//------------------------------------------------------------------------------------------

//------------------------------------------------------------------------------------------
	// 0 based indexing
	static boolean KthBitSet(int n, int k) {
		int mask = 1;
		mask = mask <<= k;
		if ((mask & n) != 0)
			return true;
		else
			return false;
	}

//------------------------------------------------------------------------------------------

	// EXTENDED EUCLIDEAN THEOREM
	// TO REPRESENT GCD IN TERMS OF A AND B
	// gcd(a,b) = a.x + b.y where x and y are integers

	static long x = -1;
	static long y = -1;

	static long gcdxy(long a, long b) {
		if (b == 0) {
			x = 1;
			y = 0;
			return a;
		} else {
			long d = gcdxy(b, a % b);
			long x1 = y;
			long y1 = x - (a / b) * y;
			x = x1;
			y = y1;

			return d;
		}
	}

//-------------------------------------------------------------------------------------------------F
}
