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;
const int maxn = 2e5 + 50;
char s[1000];
int n, a[1000], cnt[1000], vis[1000], A, B;
long long l, r;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (int i = 1; i < 101; i++) {
    if (cnt[i] == 1) {
      A++;
    }
  }
  B = 0;
  for (int i = 0; i < n; i++) {
    s[i] = 'A';
  }
  for (int i = 0; i < n; i++) {
    if (cnt[a[i]] == 1 && A - B >= 2 && !vis[a[i]]) {
      A--;
      B++;
      vis[a[i]] = 1;
      s[i] = 'B';
    }
  }
  for (int i = 0; i < n; i++) {
    if (cnt[a[i]] >= 3 && A > B && !vis[a[i]]) {
      B++;
      vis[a[i]] = 1;
      s[i] = 'B';
    }
  }
  if (A != B) {
    puts("NO");
    return 0;
  }
  puts("YES");
  for (int i = 0; i < n; i++) {
    printf("%c", s[i]);
  }
}
