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;
map<int, int> ma;
int a[205];
set<int> s;
set<int> s1;
int main() {
  int n;
  scanf("%d", &n);
  ma.clear();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ma[a[i]]++;
  }
  int v = 0;
  for (int i = 1; i <= n; i++)
    if (ma[a[i]] >= 3)
      s.insert(a[i]);
    else if (ma[a[i]] == 2)
      s1.insert(a[i]);
    else if (ma[a[i]] == 1)
      v++;
  string mm;
  if (s.size()) {
    puts("YES");
    int state = 1;
    if (v % 2 == 0) {
      for (int i = 1; i <= n; i++)
        if (s.count(a[i]) || s1.count(a[i]))
          printf("A");
        else {
          if (state)
            printf("A"), state = 0;
          else
            printf("B"), state = 1;
        }
      printf("\n");
    } else {
      bool flag = 1, label = 1;
      for (int i = 1; i <= n; i++)
        if (s.count(a[i])) {
          if (label) {
            mm.push_back('A');
            label = 0;
          } else
            mm.push_back('B');
        } else if (s1.count(a[i]))
          mm.push_back('A');
        else {
          if (state)
            mm.push_back('B'), state = 0;
          else
            mm.push_back('A'), state = 1;
        }
    }
    cout << mm << endl;
    return 0;
  } else if (v % 2 == 0) {
    puts("YES");
    bool state = 1;
    for (int i = 1; i <= n; i++)
      if (s1.count(a[i]))
        printf("A");
      else {
        if (state)
          printf("A"), state = 0;
        else
          printf("B"), state = 1;
      }
    printf("\n");
  } else
    puts("NO");
  return 0;
}
