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;
char s[105];
int t[105], a[105], tt[4];
vector<int> to[105];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    t[a[i]]++;
    to[a[i]].push_back(i);
  }
  for (int i = 1; i <= 100; ++i)
    if (t[i] > 2)
      tt[3]++;
    else
      tt[t[i]]++;
  if (tt[1] & 1 && !tt[3])
    printf("NO");
  else {
    printf("YES\n");
    for (int i = 1, j = 'A'; i <= 100; ++i)
      if (t[i] == 1) {
        s[to[i][0]] = j;
        if (j == 'A')
          j = 'B';
        else
          j = 'A';
      }
    if (tt[1] & 1)
      for (int i = 1; i <= 100; ++i)
        if (t[i] > 2) {
          s[to[i][0]] = 'B';
          break;
        }
    for (int i = 1; i <= n; ++i)
      if (!s[i]) s[i] = 'A';
    printf("%s", s + 1);
  }
  return 0;
}
