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 main() {
  long long t, i, j, n, m, k, q;
  {
    int b[1000] = {0};
    long long a[1000] = {0};
    string z;
    for (j = 0; j < 200; j++) {
      z += 'z';
      b[j] = 0;
    }
    long long count = 0;
    cin >> n;
    long long ans_i = -1;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      b[a[i]]++;
      if (b[a[i]] > 2) {
        ans_i = i;
      }
    }
    long long count_a = 0, count_b = 0;
    bool flag = true, flag2 = true;
    for (int i = 0; i < n; i++) {
      if (b[a[i]] == 1) {
        if (flag) {
          count_a++;
          z[i] = 'A';
          flag = false;
        } else if (!flag) {
          count_b++;
          z[i] = 'B';
          flag = true;
        }
      }
    }
    if (count_a != count_b) {
      if (ans_i != -1) {
        z[ans_i] = 'B';
      } else {
        flag2 = false;
      }
    }
    if (flag2) {
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        if (z[i] == 'z') {
          cout << 'A';
        } else {
          cout << z[i];
        }
      }
      cout << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}