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;
struct node {
  int id, num;
} b[150];
int a[150], w1[150], w2[150];
int main() {
  int n, i, x1, x2, k;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]].id = i;
    b[a[i]].num++;
  }
  x1 = 0;
  x2 = 0;
  for (i = 1; i <= 100; i++) {
    if (!b[i].num) continue;
    if (b[i].num == 1) {
      x1++;
      w1[x1] = b[i].id;
    } else if (b[i].num >= 3) {
      x2++;
      w2[x2] = b[i].id;
    }
  }
  sort(w1 + 1, w1 + 1 + x1);
  if (x1 % 2 == 0 || x2 >= 1) {
    printf("YES\n");
    k = 1;
    for (i = 1; i <= n; i++)
      if (w1[k] == i) {
        if (k % 2 == 1)
          printf("B");
        else
          printf("A");
        k++;
      } else if (x1 % 2 != 0 && w2[1] == i)
        printf("A");
      else
        printf("B");
  } else
    printf("NO\n");
  return 0;
}
