A string b is a subsequence of a string a if b can be obtained from a by deletion of several (possibly, zero or all) characters. For example, "xy" is a subsequence of "xzyw" and "xy", but not "yx".

You are given a string a. Your task is to reorder the characters of a so that "trygub" is not a subsequence of the resulting string.

In other words, you should find a string b which is a permutation of symbols of the string a and "trygub" is not a subsequence of b.

We have a truly marvelous proof that any string can be arranged not to contain "trygub" as a subsequence, but this problem statement is too short to contain it.

Input

The first line contains a single integer t (1≤ t≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 200) — the length of a.

The next line contains the string a of length n, consisting of lowercase English letters.

Output

For each test case, output a string b of length n which is a permutation of characters of the string a, and such that "trygub" is not a subsequence of it.

If there exist multiple possible strings b, you can print any.

Example

Input


3
11
antontrygub
15
bestcoordinator
19
trywatchinggurabruh


Output


bugyrtnotna
bestcoordinator
bruhtrywatchinggura

Note

In the first test case, "bugyrtnotna" does not contain "trygub" as a subsequence. It does contain the letters of "trygub", but not in the correct order, so it is not a subsequence.

In the second test case, we did not change the order of characters because it is not needed.

In the third test case, "bruhtrywatchinggura" does contain "trygu" as a subsequence, but not "trygub".
import java.util.*;
import java.lang.*;
import java.io.*;

public class temp1{
    
	public static void main (String[] args)throws IOException{
		
		//taking input using BufferedReader class
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		//taking total count of testcases
		int t = Integer.parseInt(br.readLine());
		while(t-->0)
		{
		    //Reading the two Strings
		    int n = Integer.parseInt(br.readLine());
		    String s = br.readLine();	
		    char[] str = new char[n];
		    for(int i=0;i<n;i++){
		    	s.getChars(0,n, str,0);
		    }
		    fun(str, n);
		}
	}
	public static void fun(char[] str, int n){
		// char[] temp={'t','r','y','g','u','b'};
		int t = 0;
		int j=n-1;
		int i =0;
		while(i<j){
			if(str[i]=='t'){
				str[i]=str[j];
				str[j]='t';
				j--;
			}
			else	i++;
		}
		// for(char st: str){
		// 	System.out.print(st);
		// }
		System.out.println(str);
	}
}// } Driver Cod