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 dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int n, a[110], frq[110];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    frq[a[i]]++;
  }
  int cnt = 0, cn = 0;
  for (int i = 0; i < (101); i++) {
    if (frq[i] && frq[i] >= 3) cn = 1;
    if (frq[i] == 1) {
      cnt++;
    }
  }
  if (cnt % 2 == 0) {
    int k = 1;
    cout << "YES" << endl;
    for (int i = 0; i < (n); i++) {
      if (frq[a[i]] == 1) {
        if (k * 2 <= cnt) {
          cout << "A";
          k++;
        } else
          cout << "B";
      } else
        cout << "B";
    }
  } else if (cn) {
    cout << "YES" << endl;
    cnt++;
    int k = 1, flg = 1;
    for (int i = 0; i < (n); i++) {
      if (frq[a[i]] == 1) {
        if (k * 2 <= cnt) {
          cout << "B";
          k++;
        } else
          cout << "A";
      } else if (flg && frq[a[i]] >= 3) {
        cout << "A";
        flg = 0;
      } else
        cout << "B";
    }
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
