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 a[105], aa[105];
int main(void) {
  int i, j, n, f, ff, z, fff, B, ok, C, pos, ans;
  scanf("%d", &n);
  memset(a, 0, sizeof(a));
  for (i = 0; i < n; i++) {
    scanf("%d", &aa[i]);
    a[aa[i]]++;
  }
  f = ff = fff = 0;
  for (i = 0; i < 105; i++) {
    if (a[i]) {
      if (a[i] == 1)
        f++;
      else if (a[i] > 2) {
        ff++;
      }
    }
  }
  for (i = n; i; i--) {
    if (a[aa[i]] > 2) {
      pos = i;
      break;
    }
  }
  if (f % 2 == 0) {
    printf("YES\n");
    ans = f / 2;
    for (i = 0; i < n; i++) {
      if (a[aa[i]] == 1 && ans) {
        printf("A");
        ans--;
      } else if (a[aa[i]] == 1) {
        printf("B");
      } else {
        printf("A");
      }
    }
    printf("\n");
  } else if (ff) {
    printf("YES\n");
    ans = f / 2;
    for (i = 0; i < n; i++) {
      if (a[aa[i]] == 1 && i == pos) {
        printf("A");
      } else if (a[aa[i]] == 1 && ans) {
        printf("A");
        ans--;
      } else if (a[aa[i]] == 1) {
        printf("B");
      } else {
        if (aa[i] == aa[pos]) {
          a[aa[i]]--;
          printf("B");
        } else
          printf("A");
      }
    }
    printf("\n");
  } else {
    printf("NO\n");
  }
}