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 dd[1001] = {};
char sig[1001] = {};
int main() {
  int n, a[1001] = {};
  memset(sig, 'A', 1001);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    dd[a[i]]++;
  }
  int num = 0;
  int renum = 0;
  for (int i = 1; i <= 101; i++) {
    if (dd[i] == 1) num++;
    if (dd[i] >= 3) renum++;
  }
  if (num % 2 == 1 && renum == 0) {
    cout << "NO";
    return 0;
  };
  int d = 1;
  if (num % 2 == 0) {
    for (int i = 1; i <= n; i++) {
      if (dd[a[i]] == 1 && d == 1) {
        sig[i] = 'B';
        d = 2;
        continue;
      }
      if (dd[a[i]] == 1 && d == 2) d = 1;
    }
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << sig[i];
    return 0;
  }
  d = 2;
  for (int i = 1; i <= n; i++) {
    if (dd[a[i]] == 1 && d == 1) {
      sig[i] = 'B';
      d = 2;
      continue;
    }
    if (dd[a[i]] == 1 && d == 2) d = 1;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++)
    if (sig[i] == 'A' && dd[a[i]] >= 3) {
      sig[i] = 'B';
      break;
    }
  for (int i = 1; i <= n; i++) cout << sig[i];
  return 0;
}