Problem: After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.

Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters). 

Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.

To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?

Input

The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.

The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".

It is guaranteed that the sum of |a| over all testcases does not exceed 100000.

Output

For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.

Example

Input


4
ANTON
NAAN
AAAAAA
OAANTTON


Output


NNOTA
AANN
AAAAAA
TNNTAOOA

Note

For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON: 

NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON. 

Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.

For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class D { //hey heyh ey

	static int[][] ps = { {0, 1, 2, 3},
			{0, 1, 3, 2},
			{0, 2, 1, 3},
			{0, 2, 3, 1},
			{0, 3, 1, 2},
			{0, 3, 2, 1},
			{1, 0, 2, 3},
			{1, 0, 3, 2},
			{1, 2, 0, 3},
			{1, 2, 3, 0},
			{1, 3, 0, 2},
			{1, 3, 2, 0},
			{2, 0, 1, 3},
			{2, 0, 3, 1},
			{2, 1, 0, 3},
			{2, 1, 3, 0},
			{2, 3, 0, 1},
			{2, 3, 1, 0},
			{3, 0, 1, 2},
			{3, 0, 2, 1},
			{3, 1, 0, 2},
			{3, 1, 2, 0},
			{3, 2, 0, 1},
			{3, 2, 1, 0} };

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int T = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		while(T-->0) {
			char[] s = code(sc.next().toCharArray());
			int n = s.length;
			int[] c = new int[4];
			for(int i = 0; i < n; i++) {
				c[s[i]-'A']++;
			}
			long maxd = -1;
			char[] res = null;
			for(int[] arr: ps) {
				char[] t = gen(c, arr, n);
				long d = dist(s, t);
				if(d > maxd) {
					maxd = d;
					res = t;
				}
			}
			sb.append(decode(res));
			sb.append("\n");
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.println(sb.toString().trim());
		pw.flush();
	}
	static char[] code(char[] s) {
		char[] res = new char[s.length];
		for(int i = 0; i < s.length; i++) {
			if(s[i] == 'A') res[i] = 'A';
			if(s[i] == 'N') res[i] = 'B';
			if(s[i] == 'T') res[i] = 'C';
			if(s[i] == 'O') res[i] = 'D';
		}
		return res;
	}
	static char[] decode(char[] s) {
		char[] res = new char[s.length];
		for(int i = 0; i < s.length; i++) {
			if(s[i] == 'A') res[i] = 'A';
			if(s[i] == 'B') res[i] = 'N';
			if(s[i] == 'C') res[i] = 'T';
			if(s[i] == 'D') res[i] = 'O';
		}
		return res;
	}
	static long dist(char[] s, char[] t) {
		int n = s.length;
		long res = 0;
		LinkedList<Integer>[] q = new LinkedList[4];
		for(int i = 0; i < 4; i++) {
			q[i] = new LinkedList<Integer>();
		}
		for(int i = 0; i < n; i++) {
			q[s[i]-'A'].add(i);
		}
		int[] arr = new int[n];
		SegmentTree st = new SegmentTree(arr);
		for(int i = 0; i < n; i++) {
			int u = q[t[i]-'A'].removeFirst();
			int ru = u + st.rmq(u, n-1);
			st.update_point(u, 1);
			res += ru-i;
		}
		return res;
	}
	static char[] gen(int[] c, int[] p, int n) {
		char[] res = new char[n];
		int ind = 0;
		for(int k = 0; k < 4; k++) {
			for(int i = 0; i < c[p[k]]; i++) {
				res[ind++] = (char)(p[k]+'A');
			}
		}
		return res;
	}
	
	static int[] next(int[] arr) {
		int n = arr.length;
		Multiset vals = new Multiset();
		for(int i = n-1; i >= 1; i--) {
			vals.add(arr[i]);
			if(arr[i] > arr[i-1]) {
				vals.add(arr[i-1]);
				arr[i-1] = vals.map.higherKey(arr[i-1]);
				vals.remove(arr[i-1]);
				for(int j = i; j < n; j++){
					arr[j] = vals.map.firstKey();
					vals.remove(arr[j]);
				}
				return arr;
			}
		}
		return null;
	} // bring in the multiset too !!!
	static class Multiset{
		TreeMap<Integer, Integer> map;
		public Multiset() {
			map = new TreeMap<>();
		}
		public void add(int a) {
			if(!map.containsKey(a)) map.put(a, 1);
			else map.put(a, map.get(a)+1);
		}
		public void remove(int a) {
			map.put(a, map.get(a)-1);
			if(map.get(a) == 0) map.remove(a);
		}
		public boolean contains(int a) {
			return map.containsKey(a);
		}
		public int count(int a) {
			if(!map.containsKey(a)) return 0;
			else return map.get(a);
		}
		public boolean isEmpty() {
			return map.isEmpty();
		}
		public boolean equals(Multiset ms) {
			for(int k: map.keySet()) {
				if(count(k) != ms.count(k)) return false;
			}
			return true;
		}
		public String toString() {
			return map.toString();
		}
	}

	static class SegmentTree {
		private int[] st; private int[] A;
		private int n;
		private int left (int p) { return p << 1; }
		private int right(int p) { return (p << 1) + 1; }
		
//		static int neutral = Integer.MAX_VALUE;					//MIN
//		static int op(int x, int y) { return Math.min(x, y); }
//		static int neutral = Integer.MIN_VALUE;					//MAX
//		static int op(int x, int y) { return Math.max(x, y); }
		static int neutral = 0;									//SUM
		static int op(int x, int y) { return x + y; }

		private void build(int p, int L, int R) {
			if (L == R)
				st[p] = A[L];
			else {
				build(left(p) , L              , (L + R) / 2);
				build(right(p), (L + R) / 2 + 1, R          );
				int v1 = st[left(p)], v2 = st[right(p)];
				st[p] = op(v1, v2);
			} }

		private int rmq(int p, int L, int R, int i, int j) {
			if (i >  R || j <  L) return neutral;
			if (L >= i && R <= j) return st[p];

			int v1 = rmq(left(p) , L              , (L+R) / 2, i, j);
			int v2 = rmq(right(p), (L+R) / 2 + 1, R          , i, j);

			return op(v1, v2);
		}

		private int update_point(int p, int L, int R, int idx, int new_value) {
			int i = idx, j = idx;
			if (i > R || j < L)
				return st[p];

			if (L == i && R == j) {
				A[i] = new_value;													// += IF DELTA
				return st[p] = A[L];
			}

			int v1, v2;
			v1 = update_point(left(p) , L              , (L + R) / 2, idx, new_value);
			v2 = update_point(right(p), (L + R) / 2 + 1, R          , idx, new_value);

			return st[p] = op(v1, v2);
		}

		public SegmentTree(int[] _A) {
			A = _A; n = A.length;
			st = new int[4 * n];
			for (int i = 0; i < 4 * n; i++) st[i] = 0;
			build(1, 0, n - 1);
		}

		public int rmq(int i, int j) { return rmq(1, 0, n - 1, i, j); }

		public int update_point(int idx, int new_value) {
			return update_point(1, 0, n - 1, idx, new_value); }
		
		public String toString() {
			return Arrays.toString(A);
		}
	}
	
	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch(IOException e) {
				throw new RuntimeException(e);
			}
		}
	}

}