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 b[100];
int main() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[a[i]]++;
  }
  int f = 0, h = -1;
  for (int i = 1; i <= 100; i++) {
    if (b[i] == 1) {
      f++;
    } else if (b[i] > 2) {
      h = i;
    }
  }
  if (f % 2 == 0) {
    cout << "YES\n";
    int r = f / 2;
    for (int i = 0; i < n; i++) {
      if (b[a[i]] == 1) {
        if (r > 0) {
          cout << "A";
          r--;
        } else {
          cout << "B";
        }
      } else {
        cout << "A";
      }
    }
  } else {
    if (h == -1) {
      return cout << "NO", 0;
    }
    int y = 0;
    cout << "YES\n";
    int r = f / 2;
    for (int i = 0; i < n; i++) {
      if ((a[i] != h) || (a[i] == h && y != 0)) {
        if (b[a[i]] == 1) {
          if (r > 0) {
            cout << "A";
            r--;
          } else {
            cout << "B";
          }
        } else {
          cout << "B";
        }
      } else {
        cout << "A";
        y++;
      }
    }
  }
}