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.*;

 
 
public class Codeforces {
	
	static int mod=1000000007 ;

	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		PrintWriter out=new PrintWriter(System.out);
	    FastScanner fs=new FastScanner();
	    int t=fs.nextInt();
	    while(t-->0) {
	    	int n=fs.nextInt();
	    	if(n<=26) {
	    		for(int i=0;i<n;i++) {
	    			out.print((char)('a'+i));
	    		}
	    		out.println();
	    	}
	    	else {
	    		int size=25;
	    		int z=n-size;
	    		if(z%2==0) {
	    			size--;
	    			z++;
	    		}
	    		int sz= z/2, ez=(z+1)/2;
	    		for(int i=0;i<sz;i++) {
	    			out.print('z');
	    		}
	    		for(int i=0;i<size;i++) {
	    			out.print((char)('a'+i));
	    		}
	    		for(int i=0;i<ez;i++) out.print('z');
	    		out.println();
	    	}
	    }
    	
	    out.close();
	}

	static long pow(long a,long b) {
		if(b<0) return 1;
		long res=1;
		while(b!=0) {
			if((b&1)!=0) {
				res*=a;
				res%=mod;
			}
			a*=a;
			a%=mod;
			b=b>>1;
		}
		return res;
	}
	static long gcd(long  a,long  b) {
		if(b==0) return a;
		return gcd(b,a%b);
	}
	static long nck(int n,int k) {
		if(k>n) return 0;
		long res=1;
		res*=fact(n);
		res%=mod;
		res*=modInv(fact(k));
		res%=mod;
		res*=modInv(fact(n-k)); 
		res%=mod;
		return res;
	}
	static long fact(long n) {
		long res=1;
		for(int i=2;i<=n;i++) {
			res*=i;
			res%=mod;
		}
		return res;
	}
	
	static long modInv(long n) {
		return pow(n,mod-2);
	}
	
	static void sort(int[] a) {
		//suffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n);
			int temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	// Use this to input code since it is faster than a Scanner
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		long[] lreadArray(int n) {
			long a[]=new long[n];
			for(int i=0;i<n;i++) a[i]=nextLong();
			return a;
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
 
}