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;
int read(int x = 0) {
  int f = 1, c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  return f * x;
}
int N, a[110], c1, c3, cnt[110], belong[110];
int main() {
  int i, res;
  N = read();
  for (i = 1; i <= N; i++) a[i] = read(), cnt[a[i]]++;
  for (i = 1; i <= 100; i++)
    if (cnt[i] == 1)
      c1++;
    else if (cnt[i] >= 3)
      c3++;
  if ((c1 & 1) == 0) {
    res = c1 >> 1;
    for (i = 1; i <= N and res; i++)
      if (cnt[a[i]] == 1) res--, belong[i] = 1;
  } else {
    if (c3 > 0) {
      res = c1 >> 1;
      for (i = 1; i <= N and res; i++)
        if (cnt[a[i]] == 1) res--, belong[i] = 1;
      res = 1;
      for (i = 1; i <= N and res; i++)
        if (cnt[a[i]] >= 3) res--, belong[i] = 1;
    } else {
      printf("NO");
      return 0;
    }
  }
  printf("YES\n");
  for (i = 1; i <= N; i++)
    if (belong[i])
      putchar('A');
    else
      putchar('B');
  return 0;
}