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;
const int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
int vetor[110];
int n, aux, i, flag;
int problemas;
int ind, solucao;
vector<int> v;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int ate_la;
  memset(vetor, 0, sizeof(vetor));
  problemas = 0;
  solucao = 0;
  flag = 1;
  for (int i = 0; i < n; i++) {
    cin >> aux;
    v.push_back(aux);
    vetor[aux]++;
    if (vetor[aux] > 2) {
      ind = i;
      solucao = 1;
    }
  }
  for (int i = 0; i < n; i++) {
    if (vetor[v[i]] == 1) {
      problemas++;
    } else if (vetor[v[i]] > 2) {
      solucao = 1;
      ate_la = problemas;
      ind = i;
    }
  }
  if (problemas % 2 && !solucao) {
    cout << "NO" << endl;
    return 0;
  }
  if (problemas % 2) {
    v[ind] = 0;
    vetor[0] = 1;
    problemas++;
    if (ate_la < problemas / 2) {
      flag = 0;
    }
  }
  cout << "YES" << endl;
  problemas /= 2;
  for (int i = 0; i < n; i++) {
    if (vetor[v[i]] == 1) {
      if (problemas) {
        cout << "A";
        problemas--;
      } else {
        cout << "B";
      }
    } else {
      if (!flag)
        cout << "B";
      else
        cout << "A";
    }
  }
  cout << endl;
}