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:
/**
 * https://codeforces.com/problemset/problem/1051/C
 * #implementation

 cnt[x]
cntNice = 0

for x in a:
    if (!cnt.containsKey(x)):
        cntNice++;
        cnt.set(x, 1);
    else:
        temp = cnt.get(x)
        cnt.set(x, temp + 1)
        if temp==1:
            cntNice--;
needNewNice = false
if cntNice % 2 == 1:
    needNewNice = true
    
    flag = false
    for key in cnt:
        if cnt.get(key) > 2:
            flag = true
            break
    if !flag:
        return FALSE
subset = 'A'
for x in a:
    if cnt.get(x) == 1:
        print subset
        subset = 'A' + 'B' - subset
    else:
        if needNewNice && cnt.get(x) > 2:
            print 'B'
            needNewNice = false
        else:
            print 'A'
 */
import java.util.Scanner;
import java.util.HashMap;

public class VasyaAndMultisets {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        HashMap<Integer, Integer> map = new HashMap<>();
        int niceNumbers = 0;
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
            int k = arr[i];
            if (map.containsKey(k)) {
                int v = map.get(k);
                map.replace(k, v + 1);
                if (v == 1) {
                    niceNumbers -= 1;
                }
            }
            else {
                niceNumbers += 1;
                map.put(k, 1);
            }
        }

        // check need to create newNiceNumber
        boolean needNiceNumber = false;
        if (niceNumbers % 2 == 1) {
            needNiceNumber = true;
            boolean canMakeNiceNumber = false;
            for (int k : map.keySet()) {
                if (map.get(k) > 2) {
                    canMakeNiceNumber = true;
                    break;
                }
            }
            // can not make nice number
            if (!canMakeNiceNumber) {
                System.out.println("NO");
                return;
            }
        }

        // print result
        System.out.println("YES");
        char subSet = 'A';
        for (int k : arr) {
            if (map.get(k) == 1) {
                System.out.print(subSet);
                // change set
                subSet = (char)('A' + 'B' - subSet);
            }
            else {
                if (needNiceNumber && map.get(k) > 2) {
                    // ood => always in subset B
                    System.out.print('B');
                    needNiceNumber = false;
                }
                else {
                    System.out.print('A');
                }
            }
        }
    }
}