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;
const int inf = 0x3f3f3f3f;
int main() {
  int n;
  scanf("%d", &n);
  int s[105], mark[105], mark1[105], mark2[105], a = 0, b = 0, c = 0;
  char ans[105];
  memset(mark, 0, sizeof(mark));
  memset(mark1, 0, sizeof(mark1));
  memset(mark2, 0, sizeof(mark2));
  for (int i = 0; i < n; i++) {
    scanf("%d", &s[i]);
    mark[s[i]]++;
    ans[i] = '1';
  }
  for (int i = 1; i <= 100; i++) {
    if (mark[i] == 1) {
      mark1[a++] = i;
    } else if (mark[i] == 2) {
      b++;
    } else if (mark[i] >= 3) {
      mark2[c++] = i;
    }
  }
  if (a & 1) {
    if (c) {
      printf("YES\n");
      int a1 = a;
      for (int i = 0; i < a / 2; i++) {
        int temp = mark1[i];
        for (int j = 0; j < n; j++) {
          if (s[j] == temp) {
            ans[j] = 'A';
            break;
          }
        }
      }
      int c1 = c, temp1 = mark2[c1 - 1];
      c1--;
      int id = 0;
      for (int i = 0; i < n; i++) {
        if (s[i] == temp1 && !id) {
          ans[i] = 'A';
          break;
        }
      }
      for (int i = 0; i < n; i++) {
        if (ans[i] != 'A') {
          ans[i] = 'B';
        }
      }
      ans[n] = '\0';
      printf("%s\n", ans);
    } else {
      printf("NO\n");
    }
  } else {
    printf("YES\n");
    int a1 = a;
    for (int i = 0; i < a / 2; i++) {
      int temp = mark1[i];
      for (int j = 0; j < n; j++) {
        if (s[j] == temp) {
          ans[j] = 'A';
          break;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      if (ans[i] != 'A') {
        ans[i] = 'B';
      }
    }
    ans[n] = '\0';
    printf("%s\n", ans);
  }
}