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;
int32_t main() {
  long long int n;
  scanf("%lld", &n);
  vector<long long int> v(n);
  map<long long int, long long int> g;
  for (long long int i = 0; i < n; ++i) {
    scanf("%lld", &v[i]);
    g[v[i]]++;
  }
  long long int c = 0;
  for (auto j = g.begin(); j != g.end(); ++j) {
    if (j->second == 1) {
      c++;
    }
  }
  if (c % 2 == 0) {
    printf("YES\n");
    char turn = 'A';
    for (long long int i = 0; i < n; ++i) {
      if (g[v[i]] == 1) {
        printf("%c", turn);
        if (turn == 'A') {
          turn = 'B';
        } else {
          turn = 'A';
        }
      } else {
        printf("A");
      }
    }
    return 0;
  }
  vector<char> ans(n);
  for (long long int i = 0; i < n; ++i) {
    ans[i] = 'N';
  }
  char turn = 'A';
  for (long long int i = 0; i < n; ++i) {
    if (g[v[i]] == 1) {
      ans[i] = turn;
      if (turn == 'A') {
        turn = 'B';
      } else {
        turn = 'A';
      }
    }
  }
  long long int y = 0;
  for (auto i = g.begin(); i != g.end(); ++i) {
    long long int x = 0;
    if (i->second > 2) {
      for (long long int j = 0; j < n; ++j) {
        if (v[j] == i->first) {
          ans[j] = turn;
          if (turn == 'A') {
            turn = 'B';
          } else {
            turn = 'A';
          }
          x = 1;
          break;
        }
      }
      if (x == 1) {
        y = 1;
        break;
      }
    }
  }
  if (y == 1) {
    printf("YES\n");
    for (long long int i = 0; i < n; ++i) {
      if (ans[i] == 'N') {
        printf("%c", turn);
      } else {
        printf("%c", ans[i]);
      }
    }
    return 0;
  }
  printf("NO");
  return 0;
}