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 = 1002;
int n, a[N];
vector<int> ind[N];
char str[N];
int main() {
  scanf("%d", &n);
  for (int(i) = (1); (i) <= (int)(n); (i)++)
    scanf("%d", &a[i]), ind[a[i]].push_back(i);
  int b = 0;
  for (int(i) = (1); (i) <= (int)(N - 1); (i)++) {
    if ((int)(ind[i]).size() == 1) {
      if (b > 0)
        str[ind[i][0]] = 'B', --b;
      else
        str[ind[i][0]] = 'A', ++b;
    }
    if ((int)(ind[i]).size() == 2) {
      str[ind[i][0]] = 'A', str[ind[i][1]] = 'B';
    }
  }
  for (int(i) = (1); (i) <= (int)(N - 1); (i)++) {
    if ((int)(ind[i]).size() > 2) {
      if (!b) {
        for (int j : ind[i]) {
          str[j] = 'A';
        }
      } else if (b > 0) {
        --b, str[ind[i][0]] = 'B';
        for (int(j) = (1); (j) <= (int)((int)(ind[i]).size() - 1); (j)++)
          str[ind[i][j]] = 'A';
      } else {
        ++b, str[ind[i][0]] = 'A';
        for (int(j) = (1); (j) <= (int)((int)(ind[i]).size() - 1); (j)++)
          str[ind[i][j]] = 'A';
      }
    }
  }
  if (!b)
    printf("YES\n%s\n", str + 1);
  else
    printf("NO\n");
  return 0;
}
