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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> s(n), check(105, 0), ans(105, -1);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    check[s[i]]++;
  }
  bool lastA = false;
  int countA = 0, countB = 0;
  for (int i = 1; i < 105; i++) {
    if (check[i] == 1) {
      if (lastA) {
        ans[i] = 1;
        countA++;
      } else {
        ans[i] = 2;
        countB++;
      }
      lastA = !lastA;
    } else if (check[i] != 0)
      ans[i] = 1;
  }
  if (countA == countB) {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      if (ans[s[i]] == 1)
        cout << "A";
      else if (ans[s[i]] == 2)
        cout << "B";
      else
        cout << "A";
    }
    cout << endl;
  } else {
    if (countA > countB) {
      int rem = abs(countA - countB), i = 1;
      while (i < 105 && rem > 0) {
        if (check[i] > 2) {
          ans[i] = 3;
          rem--;
        }
        i++;
      }
      if (rem > 0)
        cout << "NO" << endl;
      else {
        cout << "YES" << endl;
        for (int i = 0; i < n; i++) {
          if (ans[s[i]] == 1)
            cout << "A";
          else if (ans[s[i]] == 2)
            cout << "B";
          else if (ans[s[i]] == 3) {
            cout << "B";
            ans[s[i]] = 1;
          }
        }
        cout << endl;
      }
    } else {
      int rem = abs(countA - countB), i = 1;
      while (i < 105 && rem > 0) {
        if (check[i] > 2) {
          ans[i] = 3;
          rem--;
        }
        i++;
      }
      if (rem > 0)
        cout << "NO" << endl;
      else {
        cout << "YES" << endl;
        for (int i = 0; i < n; i++) {
          if (ans[s[i]] == 1)
            cout << "A";
          else if (ans[s[i]] == 2)
            cout << "B";
          else if (ans[s[i]] == 3) {
            cout << "A";
            ans[s[i]] = 2;
          }
        }
        cout << endl;
      }
    }
  }
  return 0;
}
