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
#include <bits/stdc++.h>
using namespace std;
char str[1000], tmp[1000];
int val[150], num[150], que[150];
int main() {
  int t, b, c, n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &que[i]);
    val[que[i]]++;
  }
  int a = 1, mark = -1;
  for (int i = 1; i <= 100; i++) {
    if (val[i] == 1) {
      num[i] = a;
      a = 3 - a;
    }
  }
  if (a == 2) {
    for (int i = 1; i <= n; i++) {
      if (val[que[i]] > 2) {
        mark = que[i];
        break;
      }
    }
    if (mark == -1)
      puts("NO");
    else {
      puts("YES");
      for (int i = 1; i <= n; i++) {
        if (que[i] == mark) {
          printf("B");
          mark = -1;
          continue;
        }
        if (num[que[i]] == 1)
          printf("A");
        else if (num[que[i]] == 2)
          printf("B");
        else
          printf("A");
      }
    }
  } else {
    puts("YES");
    for (int i = 1; i <= n; i++) {
      if (num[que[i]] == 1)
        printf("A");
      else if (num[que[i]] == 2)
        printf("B");
      else
        printf("A");
    }
  }
  puts("");
  return 0;
}
