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 maxn = 200005;
long long int fast_pow(long long int x, long long int y) {
  long long int ret = 1;
  while (y > 0) {
    if (y % 2 == 1) {
      ret *= x;
    }
    x *= x;
    y /= 2;
  }
  return ret;
}
long long int max(long long int x, long long int y) { return x > y ? x : y; }
long long int min(long long int x, long long int y) { return x < y ? x : y; }
long long int exgcd(long long int a, long long int b) {
  if (b == 0) return a;
  return exgcd(b, a % b);
}
int a[maxn], b[maxn], c[maxn];
int main() {
  for (int i = 0; i < 200; i++) c[i] = 105;
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
  }
  if (n == 2) {
    printf("YES\n");
    printf("AB\n");
    return 0;
  }
  int ans1 = 0, ans2 = 0;
  for (int i = 1; i <= 100; i++) {
    if (b[i] == 1)
      ans1++;
    else if (b[i] >= 3)
      ans2++;
  }
  if (ans1 % 2 == 1 && ans2 == 0)
    printf("NO\n");
  else {
    printf("YES\n");
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (b[a[i]] == 1) {
        c[i] = ans % 2;
        ans++;
      }
    }
    if (ans1 % 2 == 1)
      for (int i = 1; i <= n; i++)
        if (b[a[i]] >= 3) {
          c[i] = 1;
          break;
        }
    for (int i = 1; i <= n; i++) {
      if (c[i] == 0 || c[i] == 105)
        printf("A");
      else
        printf("B");
    }
    printf("\n");
  }
  return 0;
}