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 c[105] = {0};
int p[4] = {0};
int main() {
  int n;
  cin >> n;
  int a[n], i, x = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    c[a[i]]++;
  }
  for (i = 0; i < 105; i++) {
    if (c[i] == 1)
      p[1]++;
    else if (c[i] == 2)
      p[2]++;
    else if (c[i] >= 3)
      p[3]++;
  }
  if (p[1] % 2 == 0) {
    cout << "YES" << endl;
    int s1 = p[1] / 2, s2 = 0;
    for (i = 0; i < n; i++) {
      if (c[a[i]] == 1 && s2 < s1) {
        cout << "A";
        s2++;
      } else if (c[a[i]] == 1 && s2 >= s1) {
        cout << "B";
      } else
        cout << "B";
    }
  } else if (p[3] == 0)
    cout << "NO";
  else {
    cout << "YES\n";
    int s1 = p[1] / 2, s2 = 0, s3 = 0;
    for (i = 0; i < n; i++) {
      if (c[a[i]] == 1 && s2 < s1) {
        cout << "A";
        s2++;
      } else if (c[a[i]] == 1 && s2 >= s1)
        cout << "B";
      else if (c[a[i]] == 2)
        cout << "A";
      else if (c[a[i]] >= 3) {
        if (s3 == 0) {
          cout << "A";
          s3++;
        } else
          cout << "B";
      }
    }
  }
  return 0;
}