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 a[101];
int num[101];
char res[101];
int main() {
  int n;
  memset(num, 0, sizeof(num));
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    num[a[i]]++;
  }
  bool mmp = 0;
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    if (num[a[i]] > 2) mmp = 1;
    if (num[a[i]] == 1) sum++;
  }
  if (sum & 1 && !mmp) {
    puts("NO");
    return 0;
  }
  if (sum & 1) {
    for (int i = 0; i < n; ++i) {
      if (num[a[i]] > 2) {
        a[i] = 0;
        res[i] = 'B';
        break;
      }
    }
  }
  sum >>= 1;
  for (int i = 0; i < n; ++i)
    if (a[i]) res[i] = 'A';
  for (int i = n - 1; i >= 0 && sum; --i) {
    if (num[a[i]] == 1) {
      res[i] = 'B';
      --sum;
    }
  }
  puts("YES");
  for (int i = 0; i < n; ++i) cout << res[i];
  cout << endl;
  return 0;
}
