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;
long long int n;
unordered_set<long long int> s;
unordered_map<long long int, long long int> m;
long long int arr[105];
int main() {
  long long int i, j, k, ctr = 0;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &j);
    s.insert(j);
    m[j]++;
    arr[i] = j;
  }
  unordered_set<long long int>::iterator it;
  for (it = s.begin(); it != s.end(); it++) {
    if (m[(*it)] == 1) {
      ctr++;
    }
  }
  if (!(ctr & 1)) {
    printf("YES\n");
    long long int ctr1 = 0;
    for (i = 0; i < n; i++) {
      if (m[arr[i]] == 1) {
        if (ctr1 & 1) {
          cout << "B";
          ctr1++;
        } else {
          cout << "A";
          ctr1++;
        }
      } else {
        cout << "A";
      }
    }
    cout << endl;
  } else {
    long long int ctr1 = 0;
    for (i = 0; i < n; i++) {
      if (m[arr[i]] > 2) {
        ctr1++;
        k = arr[i];
      }
    }
    if (!ctr1) {
      printf("NO\n");
      return 0;
    }
    printf("YES\n");
    ctr1 = 0;
    for (i = 0; i < n; i++) {
      if (m[arr[i]] == 1) {
        if (ctr1 & 1) {
          cout << "B";
          ctr1++;
        } else {
          cout << "A";
          ctr1++;
        }
      } else if (arr[i] == k) {
        cout << "B";
        k = -1;
      } else {
        cout << "A";
      }
    }
    cout << endl;
  }
  return 0;
}