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 n, cnt;
int s[105], a[105];
char ans[105];
bool _free = 0;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), s[a[i]]++;
  for (int i = 1; i <= 100; i++) {
    if (s[i] == 1) cnt++;
    if (s[i] != 0 && s[i] != 1 && s[i] != 2) _free = 1;
  }
  if (cnt % 2 == 1 && (!_free))
    printf("NO\n");
  else {
    bool t = 1;
    if (cnt % 2 == 0) {
      for (int i = 1; i <= n; i++) {
        if (s[a[i]] == 1)
          ans[i] = t ? 'A' : 'B', t = !t, s[a[i]]--;
        else
          ans[i] = 'A';
      }
    } else {
      for (int i = 1; i <= n; i++)
        if (s[a[i]] == 1) ans[i] = t ? 'A' : 'B', t = !t, s[a[i]]--;
      for (int i = 1; i <= n; i++) {
        if (s[a[i]] > 2) {
          int c = a[i];
          ans[i] = t ? 'A' : 'B', t = !t;
          for (int j = i + 1; j <= n; j++)
            if (a[j] == c) ans[j] = t ? 'A' : 'B';
          break;
        }
      }
      for (int i = 1; i <= n; i++)
        if (ans[i] == 0) ans[i] = 'A';
    }
    ans[n + 1] = '\0';
    printf("YES\n%s\n", ans + 1);
  }
  return 0;
}
