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 long long inf = 707406378;
const int maxn = 200005 * 4 + 3;
void read(long long &x) {
  char ch;
  bool flag = 0;
  for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1);
       ch = getchar())
    ;
  for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar())
    ;
  x *= 1 - 2 * flag;
}
int val[maxn];
map<int, int> mp;
char ans[maxn];
int vis[maxn];
int main() {
  int n, one = 0, bigone = 0;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> val[i], mp[val[i]]++;
  for (int i = 0; i < n; i++) {
    if (mp[val[i]] == 1)
      one++;
    else if (mp[val[i]] > 2)
      bigone++;
  }
  if (one % 2 && bigone == 0) return 0 * printf("NO");
  if (one % 2 == 0) {
    int flag = 1;
    for (int i = 0; i < n; i++) {
      if (mp[val[i]] == 1) {
        if (flag)
          ans[i] = 'A';
        else
          ans[i] = 'B';
        flag = 1 - flag;
      } else
        ans[i] = 'A';
    }
  } else {
    int tmp = 0;
    int flag = 1;
    for (int i = 0; i < n; i++) {
      if (mp[val[i]] == 1) {
        if (flag)
          ans[i] = 'A';
        else
          ans[i] = 'B';
        flag = 1 - flag;
      } else {
        if (mp[val[i]] == 2) {
          ans[i] = 'A';
          continue;
        }
        if (tmp && !vis[val[i]]) {
          ans[i] = 'A';
          continue;
        }
        tmp = 1;
        if (vis[val[i]] == 0)
          ans[i] = 'B';
        else
          ans[i] = 'A';
        vis[val[i]] = 1;
      }
    }
  }
  puts("YES");
  puts(ans);
  return 0;
}