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;
const long long int M = (int)1e9 + 7;
int main() {
  int n;
  cin >> n;
  vector<int> s(n);
  vector<char> arr(n, 'A');
  vector<int> f(105, 0);
  vector<int> freq;
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    ++f[s[i]];
  }
  int count1 = 0, count2 = -1;
  for (int i = 0; i < n; ++i) {
    if (f[s[i]] == 1) {
      ++count1;
      freq.push_back(i);
    } else if (f[s[i]] >= 3) {
      count2 = i;
    }
  }
  if ((count1) % 2 != 0 && count2 == -1) {
    cout << "NO"
         << "\n";
  } else if (count1 % 2 == 0) {
    cout << "YES"
         << "\n";
    int x = count1 / 2;
    int i;
    for (i = 0; i < x; ++i) {
      arr[freq[i]] = 'A';
    }
    for (i = x; i < count1; ++i) {
      arr[freq[i]] = 'B';
    }
    for (int i = 0; i < n; ++i) {
      cout << arr[i];
    }
    cout << "\n";
  } else if (count1 % 2 != 0 && count2 != -1) {
    cout << "YES"
         << "\n";
    int x = (count1 + 1) / 2;
    int i;
    for (i = 0; i < x; ++i) {
      arr[freq[i]] = 'A';
    }
    for (i = x; i < count1; ++i) {
      arr[freq[i]] = 'B';
    }
    arr[count2] = 'B';
    for (int i = 0; i < n; ++i) {
      cout << arr[i];
    }
    cout << "\n";
  }
}
