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;
int cnt[1000] = {0}, cnt1[1000] = {0};
int main() {
  int n, a[1000], nn = 1, i, nice = 0, t = 0, cna = 0, cnb = 0, flag = 0,
                  cn = 0;
  char c[1000];
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (i = 1; i <= n; i++) {
    if (cnt[a[i]] == 1) {
      nice++;
    }
    if (cnt[a[i]] > 2) {
      cn++;
    }
  }
  if (nice % 2 == 0) {
    for (i = 1; i <= n; i++) {
      if (cnt[a[i]] == 1) {
        if (t == 0) {
          c[i] = 'A';
          t = 1;
        } else {
          c[i] = 'B';
          t = 0;
        }
      } else {
        c[i] = 'A';
      }
    }
  } else {
    if (cn == 0) {
      cout << "NO";
      return 0;
    } else {
      for (i = 1; i <= n; i++) {
        if (cnt[a[i]] == 1) {
          if (t == 0) {
            c[i] = 'A';
            t = 1;
          } else {
            c[i] = 'B';
            t = 0;
          }
        } else {
          if (flag == 0 && cnt[a[i]] > 2) {
            c[i] = 'B';
            flag = 1;
          } else
            c[i] = 'A';
        }
      }
    }
  }
  cout << "YES" << endl;
  for (i = 1; i <= n; i++) {
    cout << c[i];
  }
  return 0;
}