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 = 1e3 + 5;
int p1 = 1e9 + 7, p2 = 1e9 + 9;
int f[101];
char ans[101];
int main() {
  int n;
  scanf("%d", &n);
  int a[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    ans[i] = 'A';
    ++f[a[i]];
  }
  int o = 0, t = 0, th = 0;
  for (int i = 1; i <= 100; ++i) {
    if (f[i] == 1) ++o;
    if (f[i] == 2) ++t;
    if (f[i] >= 3) ++th;
  }
  int aa = 0, b = 0, c = 1;
  for (int i = 0; i < n; i++) {
    if (f[a[i]] == 1) {
      if (c & 1)
        ans[i] = 'A', ++aa;
      else
        ans[i] = 'B', ++b;
      c = !c;
      f[a[i]] = 0;
    }
  }
  if (aa != b && th == 0) return cout << "NO", 0;
  puts("YES");
  if (aa != b)
    for (int i = 0; i < n; i++) {
      if (f[a[i]] >= 3) {
        ans[i] = 'B';
        break;
      }
    }
  for (int i = 0; i < n; i++) {
    printf("%c", ans[i]);
  }
  return 0;
}
