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 main() {
  int N, i, j;
  scanf("%d", &N);
  int arr[N], check[101] = {
                  0,
              };
  for (i = 0; i < N; i++) scanf("%d", &arr[i]);
  for (i = 0; i < N; i++) check[arr[i]]++;
  int one = 0, more = 0, tmp, rec = 0;
  for (i = 1; i <= 100; i++) {
    if (check[i] == 1) one++;
    if (check[i] > 2) more++;
  }
  if (more == 0 && one % 2 == 1)
    printf("NO");
  else {
    printf("YES\n");
    if (one % 2 == 0) {
      tmp = -1;
      for (i = 1; i <= 100; i++) {
        if (check[i] == 1) {
          check[i] = tmp;
          if (tmp == -1)
            tmp = -2;
          else
            tmp = -1;
        }
      }
      for (i = 0; i < N; i++) {
        if (check[arr[i]] >= 0) printf("A");
        if (check[arr[i]] == -1) printf("B");
        if (check[arr[i]] == -2) printf("A");
      }
    } else {
      tmp = -1;
      for (i = 1; i <= 100; i++) {
        if (check[i] == 1) {
          check[i] = tmp;
          if (tmp == -1)
            tmp = -2;
          else
            tmp = -1;
        }
        if (check[i] > 2 && rec == 0) rec = i;
      }
      for (i = 0; i < N; i++) {
        if (check[arr[i]] >= 0) {
          if (arr[i] == rec) {
            printf("A");
            rec = 0;
          } else
            printf("B");
        }
        if (check[arr[i]] == -1) printf("B");
        if (check[arr[i]] == -2) printf("A");
      }
    }
  }
}
