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 main() {
  long long n;
  cin >> n;
  long long a[n];
  long long fre[1001] = {0};
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    fre[a[i]]++;
  }
  long long c = 0, p = 0;
  for (long long i = 0; i < 102; i++) {
    if (fre[i] == 1)
      c++;
    else if (fre[i] > 2)
      p++;
  }
  if (c % 2 and p == 0) {
    cout << "NO";
    return 0;
  }
  char s[n];
  cout << "YES\n";
  if (c % 2) {
    char ch = 'A';
    for (long long i = 0; i < n; i++) {
      if (fre[a[i]] == 1) {
        s[i] = ch;
        if (ch == 'A')
          ch = 'B';
        else
          ch = 'A';
      }
    }
    for (long long i = 0; i < n; i++) {
      if (fre[a[i]] > 2) {
        s[i] = ch;
        if (ch == 'A')
          ch = 'B';
        else
          ch = 'A';
        for (long long j = i + 1; j < n; j++)
          if (a[j] == a[i]) s[j] = ch;
        break;
      }
    }
  } else {
    char ch = 'A';
    for (long long i = 0; i < n; i++) {
      if (fre[a[i]] == 1) {
        s[i] = ch;
        if (ch == 'A')
          ch = 'B';
        else
          ch = 'A';
      } else
        s[i] = 'B';
    }
  }
  for (long long i = 0; i < n; i++) {
    if (s[i] != 'A' and s[i] != 'B') s[i] = 'B';
    cout << s[i];
  }
}