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;
map<int, int> mp;
int vis[200];
int a[200];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    vis[a[i]]++;
    if (vis[a[i]] == 1 || vis[a[i]] > 2)
      mp[a[i]] = i;
    else
      mp[a[i]] = 0;
  }
  int num1 = 0, num2 = 0;
  for (int i = 1; i <= 100; i++) {
    if (vis[i] == 1) {
      num1++;
    } else if (vis[i] > 2)
      num2++;
  }
  if ((num1 & 1) && num2 == 0) return cout << "NO" << '\n', 0;
  cout << "YES" << '\n';
  string s(n, 'A');
  int k = 0;
  if (num1 & 1) {
    num2 = 1;
    for (int i = 1; i <= 100 && k < num2; i++) {
      if (vis[i] > 2) {
        s[mp[i] - 1] = 'B';
        vis[i] = 0;
        k++;
      }
    }
  }
  k = 0;
  for (int i = 1; i <= 100 && k < num1 / 2; i++) {
    if (vis[i] == 1) {
      s[mp[i] - 1] = 'B';
      k++;
    }
  }
  cout << s << '\n';
}
