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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int Hsh[110], cnt1, cnt2, A[110], B[110];
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    A[i] = x;
    Hsh[x]++;
    if (Hsh[x] == 1) cnt1++;
    if (Hsh[x] == 2) cnt1--;
    if (Hsh[x] == 3) cnt2++;
  }
  if (cnt1 % 2 == 1 && cnt2 == 0) {
    puts("NO");
    return 0;
  }
  int t = 0, fl = 0;
  for (int i = 1; i <= n; i++) {
    if (Hsh[A[i]] == 1) B[i] = t % 2 + 1, t++;
    if (Hsh[A[i]] == 2) B[i] = 1;
    if (Hsh[A[i]] >= 3) {
      if (fl == 0 && cnt1 % 2 == 1)
        B[i] = 2, fl = 1;
      else
        B[i] = 1;
    }
  }
  puts("YES");
  for (int i = 1; i <= n; i++)
    if (B[i] == 1)
      printf("A");
    else
      printf("B");
}