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 n, cnt[101], s[100], cnt1, cnt2, cnt3, t;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &s[i]);
    cnt[s[i]]++;
  }
  for (int i = 1; i <= 100; i++) {
    if (cnt[i] == 1)
      cnt1++;
    else if (cnt[i] == 2)
      cnt2++;
    else if (cnt[i] >= 3)
      cnt3++;
  }
  if (cnt1 == 0) {
    printf("YES\n");
    for (int i = 0; i < n; i++) printf("A");
  } else if (cnt1 % 2 == 0) {
    printf("YES\n");
    t = cnt1 / 2;
    for (int i = 0; i < n; i++) {
      if (cnt[s[i]] == 1 && t) {
        cnt[s[i]]--;
        printf("A");
        t--;
      } else {
        printf("B");
      }
    }
  } else if (cnt1 % 2 == 1) {
    if (cnt3 >= 1) {
      printf("YES\n");
      t = cnt1 / 2;
      bool fin = false;
      for (int i = 0; i < n; i++) {
        if (cnt[s[i]] == 1 && t) {
          cnt[s[i]]--;
          printf("A");
          t--;
        } else if (cnt[s[i]] >= 3 && !fin) {
          fin = true;
          printf("A");
        } else {
          printf("B");
        }
      }
    } else {
      printf("NO\n");
    }
  }
}