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, x, y, i, j, a[105], s[105], b[105];
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    s[a[i]]++;
  }
  for (i = 1, j = 0; i <= 100; i++)
    if (s[i] == 1) {
      x++;
      b[i] = j;
      j = !j;
    } else if (s[i] > 2)
      y = i;
  if (!(x & 1)) {
    puts("YES");
    for (i = 1; i <= n; i++)
      if (b[a[i]])
        putchar('A');
      else
        putchar('B');
    return 0;
  }
  if (!y) {
    puts("NO");
    return 0;
  }
  puts("YES");
  for (i = 1; i <= n; i++)
    if (a[i] == y) {
      putchar('B');
      y = 0;
    } else if (!b[a[i]])
      putchar('A');
    else
      putchar('B');
  return 0;
}
