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
#include <bits/stdc++.h>
int count[110], color[110];
std::vector<int> V1, V2, V3;
char x[110];
int main() {
  int a;
  scanf("%d", &a);
  for (int i = 1; i <= a; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= a; i++) count[x[i]]++;
  for (int i = 1; i <= 100; i++)
    if (count[i] == 1) V1.push_back(i);
  for (int i = 1; i <= 100; i++)
    if (count[i] == 2) V2.push_back(i);
  for (int i = 1; i <= 100; i++)
    if (count[i] >= 3) V3.push_back(i);
  if (V1.size() % 2 == 0) {
    for (int i = 0; i < V1.size(); i += 2) color[V1[i]] = 1;
    for (int i = 1; i < V1.size(); i += 2) color[V1[i]] = 2;
    for (int i = 0; i < V2.size(); i++) color[V2[i]] = 1;
    for (int i = 0; i < V3.size(); i++) color[V3[i]] = 1;
    printf("YES\n");
    for (int i = 1; i <= a; i++) printf("%c", color[x[i]] + 'A' - 1);
  } else {
    if (V3.size() == 0) {
      printf("NO");
      return 0;
    }
    for (int i = 0; i < V1.size(); i += 2) color[V1[i]] = 1;
    for (int i = 1; i < V1.size(); i += 2) color[V1[i]] = 2;
    for (int i = 0; i < V2.size(); i++) color[V2[i]] = 1;
    for (int i = 1; i < V3.size(); i++) color[V3[i]] = 1;
    printf("YES\n");
    int control = 0;
    for (int i = 1; i <= a; i++) {
      if (color[x[i]] == 0) {
        if (control == 0) {
          printf("B");
          control++;
        } else
          printf("A");
      } else
        printf("%c", color[x[i]] + 'A' - 1);
    }
  }
}
