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 int OO = 0x3f3f3f3f;
const double eps = (1e-10);
int n, s[105], freq[105];
char ans[105];
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin >> n;
  ans[n] = '\0';
  for (int i = 0; i < (int)(n); ++i) cin >> s[i], ++freq[s[i]];
  int diff = 0;
  for (int i = 0; i < (int)(105); ++i)
    if (freq[i] == 1) ++diff;
  for (int i = 0; i < (int)(105); ++i) {
    if (diff <= 1) break;
    if (freq[i] == 1) {
      for (int j = 0; j < (int)(n); ++j)
        if (s[j] == i) ans[j] = 'B', diff -= 2;
    }
  }
  for (int i = 0; i < (int)(105); ++i) {
    if (!diff) break;
    if (freq[i] > 2) {
      for (int j = 0; j < (int)(n); ++j)
        if (s[j] == i) {
          ans[j] = 'B', --diff;
          break;
        }
    }
  }
  if (diff)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 0; i < (int)(n); ++i)
      if (ans[i] != 'B') ans[i] = 'A';
    cout << ans << '\n';
  }
  return 0;
}