Problem: 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

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class ProblemC {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[]s = new int[n];
		int[]c = new int[101];
		for(int i = 0; i < n; i++) {
			int t = scan.nextInt();
			s[i] = t;
			c[t]++;
		}
		scan.close();
		int k1 = 0;
		int k3 = 0;
		for(int i = 1; i <= 100; i++) {
			if(c[i] == 1) {
				k1++;
			}else if(c[i] >= 3){
				k3++;
			}
		}
		if(k1 % 2 == 1 && k3 == 0) {
			System.out.println("NO");
			System.exit(0);
		}
		System.out.println("YES");
		StringBuilder sb = new StringBuilder();
		if(k1 == 0) {
			for(int i = 0; i < n; i++) {
				sb.append("A");
			}
			System.out.println(sb.toString());
			System.exit(0);
		}
		if(k1 % 2 == 0) {
			int[]a = new int[k1 / 2];
			int idx = 0;
			for(int i = 1; i <= 100; i++) {
				if(c[i] == 1) {
					a[idx] = i;
					idx++;
					if(idx == k1 / 2) {
						break;
					}
				}
			}
			for(int i = 0; i < n; i++) {
				boolean flag = true;
				for(int j = 0; j < k1 / 2; j++) {
					if(s[i] == a[j]) {
						sb.append("A");
						flag = false;
						break;
					}
				}
				if(flag) {
					sb.append("B");
				}
			}
			System.out.println(sb.toString());
		}else {
			int l = (int)Math.ceil((double)k1 / 2);
			int[]a = new int[l];
			int idx = 0;

			for(int i = 1; i <= 100; i++) {
				if(c[i] == 1) {
					a[idx] = i;
					idx++;
					if(idx == l) {
						break;
					}
				}
			}
			int b = 0;
			for(int i = 1; i <= 100; i++) {
				if(c[i] >= 3) {
					b = i;
					break;
				}
			}
			int cnt = 0;
			int cnt2 = 1;
			for(int i = 0; i < n; i++) {
				boolean flag = true;
				for(int j = 0; j < l; j++) {
					if(s[i] == a[j] && cnt2 < l) {
						sb.append("B");
						cnt2++;
						flag = false;
						break;
					}
				}
				if(cnt == 0 && s[i] == b) {
					sb.append("B");
					cnt++;
					flag = false;
				}
				if(flag) {
					sb.append("A");
				}
			}
			System.out.println(sb.toString());
		}


	}
}