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 N = 120;
int n, s[N], cnt[N], s1, s3;
char ans[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]), cnt[s[i]]++;
  for (int i = 1; i <= 100; i++)
    if (cnt[i] == 1)
      s1++;
    else if (cnt[i] > 2)
      s3++;
  for (int i = 1; i <= n; i++)
    if (cnt[s[i]] == 2) ans[i] = 'A';
  if (s1 % 2 == 0) {
    puts("YES");
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (cnt[s[i]] == 1)
        ans[i] = 'A' + flag, flag ^= 1;
      else
        ans[i] = 'A';
    for (int i = 1; i <= n; i++) printf("%c", ans[i]);
    return 0;
  }
  if (!s3) {
    puts("NO");
    return 0;
  }
  puts("YES");
  int flag = 0;
  for (int i = 1; i <= n; i++)
    if (cnt[s[i]] >= 3) ans[i] = 'B' - flag, flag = 1;
  flag = 0;
  for (int i = 1; i <= n; i++)
    if (cnt[s[i]] == 1) ans[i] = 'A' + flag, flag ^= 1;
  for (int i = 1; i <= n; i++) printf("%c", ans[i]);
}
