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>
int d[1005], vis[1005];
int main() {
  int flag = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &d[i]);
    vis[d[i]]++;
  }
  int ans = 0;
  for (int i = 1; i <= 100; i++) {
    if (vis[i] == 1) {
      ans++;
    }
    if (vis[i] >= 3) {
      flag = 1;
    }
  }
  if (!(ans & 1)) {
    puts("YES");
    int g = 0;
    for (int i = 0; i < n; i++) {
      if (vis[d[i]] == 1 && g) {
        putchar('A');
      } else {
        putchar('B');
      }
      if (vis[d[i]] == 1) {
        g ^= 1;
      }
    }
    puts("");
    return 0;
  }
  if (flag == 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  int cnt = 0, f = 0;
  for (int i = 0; i < n; i++) {
    if (cnt < (ans >> 1) && vis[d[i]] == 1) {
      putchar('B');
      cnt++;
    } else if (vis[d[i]] >= 3 && f == 0) {
      putchar('B');
      f = 1;
    } else {
      putchar('A');
    }
  }
  puts("");
  return 0;
}