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;
const long double pi = acos(-1);
const long double eps = 1e-7;
int mark[120];
int mark2[120];
int s[120];
char ans[120];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    mark[s[i]]++;
  }
  int o = 0, m = 0;
  for (int i = 1; i <= 100; i++) {
    if (mark[i] == 1) o++;
    if (mark[i] > 2) m++;
  }
  for (int i = 0; i <= 110; i++) ans[i] = 'k';
  if (o % 2 == 1 && m == 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  if (o % 2 == 1 && m > 0) {
    int flg = 0;
    for (int i = 1; i <= n; i++)
      if (mark[s[i]] == 1) {
        if (flg == 0)
          ans[i] = 'A';
        else
          ans[i] = 'B';
        flg = 1 - flg;
      }
    for (int i = 1; i <= n; i++)
      if (mark[s[i]] > 2) {
        if (flg == 0) {
          ans[i] = 'A';
          for (int j = i + 1; j <= n; j++)
            if (s[j] == s[i]) ans[j] = 'B';
        } else {
          ans[i] = 'B';
          for (int j = i + 1; j <= n; j++)
            if (s[j] == s[i]) ans[j] = 'A';
        }
        break;
      }
    for (int i = 1; i <= n; i++)
      if (ans[i] == 'k') ans[i] = 'A';
  }
  if (o % 2 == 0) {
    int flg = 0;
    for (int i = 1; i <= n; i++)
      if (mark[s[i]] == 1) {
        if (flg == 0)
          ans[i] = 'A';
        else
          ans[i] = 'B';
        flg = 1 - flg;
      }
    for (int i = 1; i <= n; i++)
      if (ans[i] == 'k') ans[i] = 'A';
  }
  for (int i = 1; i <= n; i++) cout << ans[i];
  cout << endl;
  return 0;
}