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;
vector<int> nums[105];
char ans[105];
int main(void) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    nums[tmp].push_back(i);
  }
  int ones = 0;
  int upid = -1;
  for (int i = 0; i <= 100; i++) {
    if (nums[i].size() == 1) ones++;
    if (nums[i].size() >= 3) upid = i;
  }
  if (ones % 2 == 0) {
    bool infst = false;
    for (int i = 0; i <= 100; i++) {
      if (nums[i].size() == 1) {
        if (infst) {
          ans[nums[i][0]] = 'A';
        } else {
          ans[nums[i][0]] = 'B';
        }
        infst = !infst;
      } else {
        for (auto elm : nums[i]) ans[elm] = 'A';
      }
    }
  } else {
    if (upid != -1) {
      bool infst = true;
      for (int i = 0; i <= 100; i++) {
        if (nums[i].size() == 1) {
          if (infst) {
            ans[nums[i][0]] = 'A';
          } else {
            ans[nums[i][0]] = 'B';
          }
          infst = !infst;
        } else {
          for (auto elm : nums[i]) ans[elm] = 'A';
          if (i == upid) ans[nums[i][0]] = 'B';
        }
      }
    } else {
      printf("NO");
      return 0;
    }
  }
  ans[n + 1] = 0;
  printf("YES\n%s", ans + 1);
  return 0;
}