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;
const int N = 100;
int n;
int a[N];
int amt[N];
bool to[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    --a[i];
    ++amt[a[i]];
  }
  int s = 0;
  int t = -1;
  for (int i = 0; i < N; ++i) {
    if (amt[i] == 1) {
      to[i] = s & 1;
      ++s;
    } else if (amt[i] > 2)
      t = i;
  }
  if ((s & 1) && t == -1)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 0; i < n; ++i) {
      if (amt[a[i]] == 1)
        printf("%c", 'A' + to[a[i]]);
      else if ((s & 1) && a[i] == t) {
        t = -1;
        printf("%c", 'B');
      } else
        printf("%c", 'A');
    }
    printf("\n");
  }
  return 0;
}
