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 main() {
  int n;
  cin >> n;
  int a[100];
  int freq[101] = {};
  bool targee3 = 0;
  int u = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    freq[a[i]]++;
    if (freq[a[i]] == 1)
      u++;
    else if (freq[a[i]] == 2)
      u--;
    else
      targee3 = 1;
  }
  if (u % 2) {
    if (!targee3) {
      printf("NO");
      return 0;
    }
  } else
    targee3 = false;
  bool A = true;
  printf("YES\n");
  for (int i = 0; i < n; i++) {
    if (freq[a[i]] == 1) {
      if (A)
        printf("A");
      else
        printf("B");
      A = !A;
    } else if (freq[a[i]] == 2) {
      printf("A");
    } else if (targee3) {
      printf("B");
      targee3 = false;
    } else {
      printf("A");
    }
  }
  return 0;
}
