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 main() {
  long long n, flag = 0, f = 0, t = 0;
  cin >> n;
  long long arr[n + 5], pos[1000], col[1000];
  memset(col, -1, sizeof(col));
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    pos[arr[i]] = i;
    mp[arr[i]]++;
  }
  for (auto it : mp) {
    if (it.second == 1)
      f++;
    else if (it.second >= 3)
      t++;
  }
  if (f % 2 == 1 and t == 0) {
    return cout << "NO", 0;
  }
  cout << "YES" << endl;
  for (auto it : mp) {
    long long x = it.first;
    if (it.second == 1) {
      if (flag == 0)
        col[pos[x]] = 1;
      else
        col[pos[x]] = 2;
      flag ^= 1;
    }
  }
  if (f % 2 == 0) {
    for (long long i = 1; i <= n; i++) {
      if (col[i] == -1)
        cout << "A";
      else {
        if (col[i] == 1)
          cout << "A";
        else
          cout << "B";
      }
    }
  } else if (f % 2) {
    for (auto it : mp) {
      long long x = it.first;
      if (it.second >= 3) {
        col[pos[x]] = 2;
        break;
      }
    }
    for (long long i = 1; i <= n; i++) {
      if (col[i] == -1)
        cout << "A";
      else {
        if (col[i] == 1)
          cout << "A";
        else
          cout << "B";
      }
    }
  }
  cout << endl;
}
