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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxm = 105;
int flag[maxm], a[maxm], b[maxm];
char str[maxm];
int main() {
  int n, j, k, sum, i, rev;
  sum = rev = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    flag[a[i]]++;
  }
  for (i = 1; i <= n; i++) {
    if (flag[a[i]] == 1) b[++sum] = a[i];
    if (flag[a[i]] > 2) rev++;
  }
  if (sum % 2 == 1 && rev == 0) {
    printf("NO\n");
    return 0;
  }
  int now = 1;
  for (i = 1; i <= n; i++) {
    if (flag[a[i]] == 1) {
      if (now <= sum / 2)
        str[i] = 'A';
      else
        str[i] = 'B';
      now++;
    } else if (flag[a[i]] > 2 && sum % 2 == 1)
      str[i] = 'A', sum--;
    else
      str[i] = 'B';
  }
  printf("YES\n");
  for (i = 1; i <= n; i++) printf("%c", str[i]);
  printf("\n");
  return 0;
}