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, k;
vector<int> Sin;
int numm[101], pos[101];
struct Node {
  int num;
  char grp;
  Node() { grp = 'A'; }
} a[101];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i].num), numm[a[i].num]++, pos[a[i].num] = i;
  for (int i = 1; i <= 100; i++) {
    if (numm[i] == 1)
      Sin.push_back(i);
    else if (numm[i] > 2)
      k = i;
  }
  if (Sin.size() % 2 != 0 && k == 0) return cout << "NO", 0;
  cout << "YES\n";
  for (int i = 0; i < Sin.size() / 2; i++) a[pos[Sin[i]]].grp = 'B';
  if (Sin.size() % 2 != 0) a[pos[k]].grp = 'B';
  for (int i = 1; i <= n; i++) cout << a[i].grp;
  return 0;
}