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 n, suma, sumb;
int sum[105];
bool t[105], flag;
vector<int> num[105];
int main() {
  scanf("%d", &n);
  int v;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v);
    sum[v]++;
    num[v].push_back(i);
  }
  for (int i = 1; i <= 100; i++) {
    if (sum[i] == 1) {
      suma++;
      continue;
    }
    if (sum[i] == 2) continue;
    if (sum[i] >= 3) {
      sumb++;
      continue;
    }
  }
  if (suma & 1 && !sumb) {
    printf("NO");
    return 0;
  }
  if (suma & 1) {
    suma /= 2;
    flag = true;
    for (int i = 1; i <= 100; i++) {
      if (sum[i] == 1 && suma) {
        t[num[i][0]] = true;
        suma--;
      }
      if (sum[i] >= 3 && flag) {
        t[num[i][0]] = true;
        flag = false;
      }
    }
  } else {
    suma /= 2;
    for (int i = 1; i <= 100; i++) {
      if (sum[i] == 1 && suma) {
        t[num[i][0]] = true;
        suma--;
      }
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++)
    if (t[i])
      printf("A");
    else
      printf("B");
  return 0;
}