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>
int a, i, x[102], m, o, oo, p, q, sum[102];
int main() {
  scanf("%d", &a);
  for (i = 0; i < a; i++) {
    scanf("%d", &x[i]);
    sum[x[i]]++;
  }
  for (i = 1; i <= 100; i++) {
    if (sum[i] == 1)
      p++;
    else if (sum[i] >= 3)
      m = i, q++;
  }
  if (p % 2 == 0) {
    puts("YES");
    for (i = 0; i < a; i++) {
      if (sum[x[i]] == 1) {
        if (o < p / 2)
          printf("A");
        else
          printf("B");
        o++;
      } else
        printf("A");
    }
  } else {
    if (q) {
      puts("YES");
      for (i = 0; i < a; i++) {
        if (m == x[i]) {
          if (oo == 0)
            printf("A");
          else
            printf("B");
          oo++;
        } else if (sum[x[i]] == 1) {
          if (o < p / 2)
            printf("A");
          else
            printf("B");
          o++;
        } else
          printf("A");
      }
    } else
      puts("NO");
  }
}