Problem: You are given a string, consisting of lowercase Latin letters.

A pair of neighbouring letters in a string is considered ugly if these letters are also neighbouring in a alphabet. For example, string "abaca" contains ugly pairs at positions (1, 2) — "ab" and (2, 3) — "ba". Letters 'a' and 'z' aren't considered neighbouring in a alphabet.

Can you rearrange the letters of a given string so that there are no ugly pairs? You can choose any order of the letters of the given string but you can't add any new letters or remove the existing ones. You can also leave the order the same.

If there are multiple answers, print any of them.

You also have to answer T separate queries.

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of queries.

Each of the next T lines contains string s (1 ≤ |s| ≤ 100) — the string for the next query. It is guaranteed that it contains only lowercase Latin letters.

Note that in hacks you have to set T = 1.

Output

Print T lines. The i-th line should contain the answer to the i-th query.

If the answer for the i-th query exists, then print such a rearrangment of letters of the given string that it contains no ugly pairs. You can choose any order of the letters of the given string but you can't add any new letters or remove the existing ones. You can also leave the order the same.

If there are multiple answers, print any of them.

Otherwise print "No answer" for that query.

Example

Input


4
abcd
gg
codeforces
abaca


Output


cadb
gg
codfoerces
No answer

Note

In the first example answer "bdac" is also correct.

The second example showcases the fact that only neighbouring in alphabet letters are not allowed. The same letter is ok.

There are lots of valid answers for the third example.

Solution (in JAVA):

Solution:


import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0){
			String s=sc.next();
			char ctr[]=s.toCharArray();
			int letter[]=new int[26];//存储每个字母出现的次数
			for(int i=0;i<ctr.length;i++){
				int num=ctr[i]-'a';
				letter[num]++;
			}
			char goal[]=new char[26];//把出现过的字母按从小到大的顺序存储一次
			int len=0;
			for(int i=0;i<26;i++){
				if(letter[i]!=0)
					goal[len++]=(char) (i+'a');
			}
			String str="";
			
			if(len%2==0){//分奇偶来构造
				for(int i=0;i<len/2-1;i++){
					str+=goal[i];
					str+=goal[len-i-1];
				}
				String tmp=goal[len/2]+str+goal[len/2-1];	
				char ct[]=tmp.toCharArray();
				boolean flag=true;//判断构造完之后是否没有丑数
				for(int i=0;i<ct.length-1;i++){
					int num=ct[i]-ct[i+1];
					if(num==1||num==-1){
						flag=false;
						break;
					}
				}
				if(flag){
					for(int i=0;i<ct.length;i++){
						for(int j=0;j<letter[(int)(ct[i]-'a')];j++){
							System.out.print(ct[i]);
						}
					}
					System.out.println();	
				}else{
					System.out.println("No answer");
				}
			}else{
				for(int i=0;i<len/2;i++){
					str+=goal[i];  //注意不能写成str+=goal[i]+goal[len-i-1],这样会先计算后两个的ascall码和，是整型
					str+=goal[len-i-1];
				}
				String tmp=str;
				tmp+=goal[len/2];
				char ct[]=tmp.toCharArray();
				boolean flag1=true;
				for(int i=0;i<ct.length-1;i++){
					int num=ct[i]-ct[i+1];
					if(num==1||num==-1){
						flag1=false;
						break;
					}
				}
				if(flag1){
					for(int i=0;i<ct.length;i++){
						for(int j=0;j<letter[ct[i]-'a'];j++){
							System.out.print(ct[i]);
						}
					}
					System.out.println();					
				}else{
					tmp=goal[len/2]+str;
					char ct2[]=tmp.toCharArray();
					boolean flag2=true;
					for(int i=0;i<ct.length-1;i++){
						int num=ct2[i]-ct2[i+1];
						if(num==1||num==-1){
							flag2=false;
							break;
						}
					}
					if(flag2){
						for(int i=0;i<ct2.length;i++){
							for(int j=0;j<letter[ct2[i]-'a'];j++){
								System.out.print(ct2[i]);
							}
						}
						System.out.println();	
					}else{
						System.out.println("No answer");
					}
				}
			}
		}
		sc.close();
	}
}