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;
struct NODE {
  int o, v, n;
};
NODE a[105];
int ans[105];
map<int, int> ma;
int main() {
  int n;
  scanf("%d", &n);
  int one = 0, two = 0, mo = 0;
  ma.clear();
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].v);
    a[i].o = i;
    a[i].n = 0;
    if (ma.find(a[i].v) == ma.end()) ma[a[i].v] = 0;
    ma[a[i].v]++;
  }
  for (auto i = ma.begin(); i != ma.end(); i++) {
    if (i->second == 1)
      one++;
    else if (i->second == 2)
      two++;
    else
      mo++;
  }
  bool flag = true;
  if (one % 2 == 0) {
    int p = one / 2;
    for (int i = 0; i < n; i++) {
      if (ma[a[i].v] == 1) {
        a[i].n = 1;
        p--;
      }
      if (!p) break;
    }
  } else {
    if (mo == 0)
      flag = false;
    else {
      int p = one / 2, om = 1;
      for (int i = 0; i < n; i++) {
        if (p && ma[a[i].v] == 1) {
          a[i].n = 1;
          p--;
        } else if (om && ma[a[i].v] > 2) {
          a[i].n = 1;
          om--;
        }
        if (!p && !om) break;
      }
    }
  }
  printf("%s\n", flag ? "YES" : "NO");
  if (flag) {
    for (int i = 0; i < n; i++) ans[a[i].o] = a[i].n;
    for (int i = 0; i < n; i++) printf("%s", ans[i] ? "A" : "B");
    printf("\n");
  }
  return 0;
}