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.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.*;


public class ForcesC {

	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}


	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long mod = (long)(1e9+7);
	static FastReader sc = new FastReader();
    static List<Integer>[] edge ;
	public static void main(String[] args) {
		int t = sc.nextInt();
		while(t-->0) {
			String s = sc.next();
			int[] freq = new int[26];
			int min = 27;
			for(int i=0;i<s.length();++i) {
				freq[s.charAt(i)-'a']++;
				
			}
		   // System.out.println(Arrays.toString(freq));
			
			StringBuffer sb = new StringBuffer();
			for(int i=0;i<26;i+=2) {
				while(freq[i]>0) {
					sb.append((char)(i+'a'));
					--freq[i];
				}
			}
			StringBuffer sb2 = new StringBuffer();
			for(int i=1;i<26;i+=2) {
				while(freq[i]>0) {
					sb2.append((char)(i+'a'));
					--freq[i];
				}
			}
		   StringBuffer ans;
		   if(sb.length() == 0) ans = new StringBuffer(sb2);
		   else if(sb2.length() == 0) ans = new StringBuffer(sb);
		   else if(Math.abs(sb.charAt(sb.length()-1)-sb2.charAt(0))== 1)
			   ans = new StringBuffer(sb2.append(sb));
		   else ans = new StringBuffer(sb.append(sb2));
		   boolean found = true ;
			for(int i =1;i<ans.length();i++){
				if(Math.abs(ans.charAt(i)-ans.charAt(i-1)) == 1) {
					found = false ;
					break;
				}
			}
			if(found)
				out.println(ans);
			else out.println("No answer");
			}
		
		out.close();
		
	}
}