Vasya has a multiset s consisting of n integer numbers. Vasya calls some number x nice if it appears in the multiset exactly once. For example, multiset \{1, 1, 2, 3, 3, 3, 4\} contains nice numbers 2 and 4.

Vasya wants to split multiset s into two multisets a and b (one of which may be empty) in such a way that the quantity of nice numbers in multiset a would be the same as the quantity of nice numbers in multiset b (the quantity of numbers to appear exactly once in multiset a and the quantity of numbers to appear exactly once in multiset b).

Input

The first line contains a single integer n~(2 ≤ n ≤ 100).

The second line contains n integers s_1, s_2, ... s_n~(1 ≤ s_i ≤ 100) — the multiset s.

Output

If there exists no split of s to satisfy the given requirements, then print "NO" in the first line.

Otherwise print "YES" in the first line.

The second line should contain a string, consisting of n characters. i-th character should be equal to 'A' if the i-th element of multiset s goes to multiset a and 'B' if if the i-th element of multiset s goes to multiset b. Elements are numbered from 1 to n in the order they are given in the input.

If there exist multiple solutions, then print any of them.

Examples

Input

4
3 5 7 1


Output

YES
BABA


Input

3
3 5 1


Output

NO
import java.util.*;
import java.io.*;

public class File {
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

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

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }

            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

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

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));



		
		
		int n = sc.nextInt();
		
		int[] counts = new int[101];
		int[] a = new int[n];
		
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			counts[a[i]]++;
		}
		
		
		int ones = 0;
		int twos = 0;
		int greater = 0;
		
		for (int count : counts) {
			if (count == 1) {
				ones++;
			}
			else if (count == 2) {
				twos++;
			}
			else if (count > 2) {
				greater++;
			}
		}
		
		boolean isPossible = false;
		
		// Will be evenly divided.
		if (ones % 2 == 0) {
			isPossible = true;
		}
		else {
			// One will be offset, see if we can change one extra to a 1 count.
			if (greater > 0) {
				isPossible = true;
			}
		}
		
		
		if (isPossible) {
			// Count == 1, MUST add 1 to one.
			// Count == 2, ALWAYS evens out. Add to either multiset.
			// Count > 2, count as 1, or as neither (add all to one side).
			
			// Split the 1 count elements into both groups.
			int[] assigned = new int[n];
			
			
			int half = ones / 2;
			int remHalf1 = half;
			
			for (int i = 0; i < n; i++) {
				int val = a[i];
				
				if (counts[val] == 1) {
					if (remHalf1 > 0) {
						assigned[i] = 1;
						remHalf1--;
					}
					else {
						assigned[i] = 2;
					}
				}
				// Assign the two counts to first group.
				else if (counts[val] == 2) {
					assigned[i] = 1;
				}
			}
			
			boolean hasExtra = ones % 2 == 1;
			
			// Assign 1 greater count to first group if ones was odd.
			for (int i = 0; i < n; i++) {
				int val = a[i];
				
				if (counts[val] > 2) {
					if (hasExtra) {
					    hasExtra = false;
						assigned[i] = 1;
					}
					else {
						assigned[i] = 2;
					}
				}
			}
			
			
			out.println("YES");
			for (int num : assigned) {
				if (num == 1) {
					out.print("A");
				}
				else {
					out.print("B");
				}
			}
		}
		else {
			out.println("NO");
		}
		
		
		
		

		
		
		out.close();
	}
}











	