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;
template <typename T>
void read(T &num) {
  char c = getchar();
  num = 0;
  T f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    num = (num << 3) + (num << 1) + (c ^ 48);
    c = getchar();
  }
  num *= f;
}
template <typename T>
void qwq(T x) {
  if (x > 9) qwq(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void write(T x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  qwq(x);
  putchar('\n');
}
int container[110];
multiset<int> v;
int main() {
  int n;
  read(n);
  for (int i = 1; i <= n; i++) {
    read(container[i]);
    v.insert(container[i]);
  }
  set<int>::iterator it;
  int tmp = 0;
  int chk = 0;
  int t = 0;
  for (it = v.begin(); it != v.end(); it++) {
    int nop = *it;
    tmp += (v.count(nop) == 1);
    if (v.count(nop) > 2) {
      t = nop;
      chk = 1;
    }
  }
  if (tmp % 2 == 1 && !chk) {
    puts("NO");
    return 0;
  } else {
    puts("YES");
  }
  if (tmp % 2 == 0) {
    int temp = 0;
    for (int i = 1; i <= n; i++) {
      if (v.count(container[i]) == 1) {
        temp++;
        if (temp <= tmp / 2)
          putchar('A');
        else
          putchar('B');
      } else {
        putchar('A');
      }
    }
  } else {
    int temp = 0;
    bool flag = false;
    for (int i = 1; i <= n; i++) {
      if (v.count(container[i]) == 1) {
        temp++;
        if (temp <= tmp / 2)
          putchar('A');
        else
          putchar('B');
      } else {
        if (t == container[i] && flag == false) {
          flag = true;
          putchar('A');
        } else {
          putchar('B');
        }
      }
    }
  }
  putchar('\n');
  return 0;
}