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 n, ar[1003], cnt[1003], flag, tag, num, a, x, val[1003];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    cnt[x]++;
    ar[i] = x;
  }
  for (int i = 1; i <= 100; i++) {
    if (cnt[i] == 1) {
      num++;
      if (tag) val[i] = 1;
      tag ^= 1;
    } else if (cnt[i] > 2) {
      a = i;
    }
  }
  if ((num & 1) && !a) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    if (num & 1) flag = 1;
    for (int i = 1; i <= n; i++) {
      x = ar[i];
      if (flag && x == a) {
        printf("B");
        flag = 0;
      } else {
        if (val[x])
          printf("B");
        else
          printf("A");
      }
    }
    printf("\n");
  }
  return 0;
}
