Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0), EPS = 1e-9;
const int INF = 1e9, NINF = -1e9, MOD = 1e9 + 7;
int dx[8] = {-1, -1, 0, 1, 1, 1, 0, -1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void GO() {
  cout << fixed << setprecision(20);
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
}
int a[105], f[105];
int main() {
  GO();
  int n, cnt = 0, cnt1 = 0, mA = 0, mB = 0;
  bool flag = 1;
  char s[105];
  memset(s, 74, sizeof(s));
  cin >> n;
  for (int i = 0; i < (int)n; i++) cin >> a[i], ++f[a[i]];
  for (int i = 0; i < (int)102; i++)
    if (f[i] == 1)
      ++cnt;
    else if (f[i] > 2)
      ++cnt1;
  if ((cnt & 1) && !cnt1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    if (!(cnt % 2)) {
      for (int i = 0; i < (int)n; i++)
        if (f[a[i]] == 1) {
          if (flag)
            s[i] = 'A', flag = 0;
          else
            s[i] = 'B', flag = 1;
        }
      for (int i = 0; i < (int)n; i++)
        if (s[i] == 'J') s[i] = 'A';
      for (int i = 0; i < (int)n; i++) cout << s[i];
    } else {
      for (int i = 0; i < (int)n; i++)
        if (f[a[i]] == 1) {
          if (flag)
            s[i] = 'A', flag = 0, ++mB;
          else
            s[i] = 'B', flag = 1, ++mA;
        }
      for (int i = 0; i < (int)n; i++)
        if (f[a[i]] > 2) {
          if (mB > mA)
            s[i] = 'B';
          else
            s[i] = 'A';
          break;
        }
      for (int i = 0; i < (int)n; i++)
        if (s[i] == 'J') {
          if (mB > mA)
            s[i] = 'A';
          else
            s[i] = 'B';
        }
      for (int i = 0; i < (int)n; i++) cout << s[i];
    }
  }
  return 0;
}