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 a[200];
int c[200];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    c[a[i]]++;
  }
  int s = 0;
  int pp = 0;
  for (int i = 1; i <= 100; i++) {
    if (c[i] == 1) {
      s++;
    } else if (c[i] > 2 && pp == 0) {
      pp = i;
    }
  }
  if (s % 2 == 1 && pp == 0) {
    printf("NO\n");
  } else if (s % 2 == 1 && pp > 0) {
    printf("YES\n");
    int co = 0;
    int bo = 0;
    s = (int)s / 2;
    for (int i = 1; i <= n; i++) {
      if (c[a[i]] == 1 && co < s) {
        printf("A");
        co++;
      } else if (c[a[i]] == 1) {
        printf("B");
      } else if (a[i] == pp && bo == 0) {
        printf("A");
        bo += 10;
      } else {
        printf("B");
      }
    }
    printf("\n");
  } else if (s % 2 == 0) {
    printf("YES\n");
    int co = 0;
    s = s / 2;
    for (int i = 1; i <= n; i++) {
      if (c[a[i]] == 1 && co < s) {
        printf("A");
        co++;
      } else {
        printf("B");
      }
    }
    printf("\n");
  }
}