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
#include <bits/stdc++.h>
using namespace std;
int a[105], num[105][105], v[2];
char s[105];
int main() {
  int n, i, j;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    num[a[i]][0]++;
    num[a[i]][num[a[i]][0]] = i;
  }
  int flag = 0;
  for (i = 1; i <= 100; i++) {
    if (num[i][0] == 1) {
      v[flag]++;
      num[i][0]--;
      s[num[i][1]] = 'A' + flag;
      flag = !flag;
    }
  }
  if (!flag) {
    for (i = 1; i <= 100; i++) {
      if (num[i][0]) {
        for (j = 1; j <= num[i][0]; j++) {
          s[num[i][j]] = 'A' + flag;
        }
      }
    }
    cout << "YES" << endl;
    for (i = 1; i <= n; i++) cout << s[i];
    cout << endl;
  } else {
    for (i = 1; i <= 100; i++) {
      if (num[i][0] >= 3) {
        s[num[i][num[i][0]]] = flag + 'A';
        v[flag]++;
        flag = !flag;
        num[i][0]--;
        break;
      }
    }
    for (i = 1; i <= 100; i++) {
      if (num[i][0]) {
        for (j = 1; j <= num[i][0]; j++) {
          s[num[i][j]] = 'A' + flag;
        }
      }
    }
    if (v[0] != v[1])
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      for (i = 1; i <= n; i++) cout << s[i];
      cout << endl;
    }
  }
  return 0;
}
