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;
int n;
int a[101];
int c[101];
char b[101];
int tw;
int cnt;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    c[a[i]]++;
  }
  for (int i = 1; i < 101; i++) {
    if (c[i] == 1) {
      cnt++;
    } else if (c[i] == 2) {
      tw++;
    }
  }
  if (n - cnt - tw < 3 && n % 2 == 1) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  if (cnt % 2 == 0) {
    int cur = 0;
    for (int i = 0; i < n; i++) {
      if (c[a[i]] == 1) {
        cur++;
        if (cur <= cnt / 2) {
          cout << 'A';
        } else {
          cout << 'B';
        }
      } else {
        cout << 'A';
      }
    }
  } else {
    int cur = 0;
    int ok = 0;
    for (int i = 0; i < n; i++) {
      if (c[a[i]] == 1) {
        cur++;
        if (cur <= cnt / 2) {
          cout << 'A';
        } else {
          cout << 'B';
        }
      } else if (!ok && c[a[i]] >= 3) {
        ok++;
        cout << 'A';
      } else {
        cout << 'B';
      }
    }
  }
}
