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;
int a[105], c[105], pos[105];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    c[a[i]]++;
  }
  int s = 0, flag = 1;
  for (int i = 1; i <= 100; i++)
    if (c[i] == 1) s++;
  if (s % 2 == 0) {
    int g = 0;
    for (int i = 1; g * 2 < s; i++)
      if (c[a[i]] == 1) pos[i] = 1, g++;
  } else {
    flag = 0;
    for (int i = 1; i <= 100; i++)
      if (c[i] > 2) {
        flag = i;
        break;
      }
    if (flag) {
      int g = 0;
      for (int i = 1; g < s / 2; i++)
        if (c[a[i]] == 1) pos[i] = 1, g++;
      for (int i = 1; i <= n; i++)
        if (a[i] == flag) {
          pos[i] = 1;
          break;
        }
    }
  }
  if (!flag)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%c", pos[i] ? 'A' : 'B');
    printf("\n");
  }
  return 0;
}
