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 ai[105], bi[105];
int ci[105];
int main() {
  int n;
  cin >> n;
  memset(bi, 0, sizeof(bi));
  for (int i = 1; i <= n; i++) {
    cin >> ai[i];
    bi[ai[i]]++;
  }
  int sum1 = 0, sum2 = 0, sum3 = 0;
  for (int i = 1; i <= 100; i++) {
    if (bi[i] >= 3) {
      sum1++;
    } else if (bi[i] == 1) {
      sum3++;
    } else if (bi[i] == 2) {
      sum2++;
    }
  }
  if (sum3 % 2 == 1 && sum1 != 0) {
    cout << "YES" << endl;
    int flag = 1, fla = 0;
    for (int i = 1; i <= n; i++) {
      if (ci[ai[i]] == 1 && fla == 1) {
        cout << "A";
      } else if (ci[ai[i]] == 2 && fla == 1) {
        cout << "B";
      } else if (bi[ai[i]] == 1 && flag == 1) {
        cout << "A";
        flag = 0;
      } else if (bi[ai[i]] == 1 && flag == 0) {
        cout << "B";
        flag = 1;
      } else if (bi[ai[i]] == 2) {
        cout << "A";
      } else if (bi[ai[i]] >= 3 && flag == 1 && fla == 0) {
        cout << "A";
        flag = 0;
        ci[ai[i]] = 2;
        fla = 1;
      } else if (bi[ai[i]] >= 3 && flag == 0 && fla == 0) {
        cout << "B";
        flag = 1;
        ci[ai[i]] = 1;
        fla = 1;
      } else
        cout << "A";
    }
  } else if (sum3 % 2 == 0) {
    cout << "YES" << endl;
    int flag = 1;
    for (int i = 1; i <= n; i++) {
      if (bi[ai[i]] == 1 && flag == 1) {
        cout << "A";
        flag = 0;
      } else if (bi[ai[i]] == 1 && flag == 0) {
        cout << "B";
        flag = 1;
      } else
        cout << "A";
    }
  } else {
    cout << "NO" << endl;
  }
  return 0;
}