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
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
priority_queue<int, vector<int>, greater<int> > pq;
vector<int> v;
multiset<int> ms;
int vec[100];
int num[101];
char ans[101];
vector<int> o, t, th;
queue<int> lo[101];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &vec[i]);
    lo[vec[i]].push(i);
    num[vec[i]]++;
  }
  for (int i = 1; i <= 100; i++) {
    if (num[i] == 1) o.push_back(i);
    if (num[i] == 2) t.push_back(i);
    if (num[i] >= 3) th.push_back(i);
  }
  if ((int)(o).size() % 2 && (int)(th).size() == 0) return !printf("NO");
  printf("YES\n");
  int tt;
  for (int i = 0; i < (int)(o).size() / 2; i++) {
    tt = lo[o[i]].front();
    lo[o[i]].pop();
    ans[tt] = 'A';
  }
  for (int i = (int)(o).size() / 2; i < (int)(o).size(); i++) {
    tt = lo[o[i]].front();
    lo[o[i]].pop();
    ans[tt] = 'B';
  }
  for (int i = 0; i < (int)(t).size(); i++) {
    while (!lo[t[i]].empty()) {
      tt = lo[t[i]].front();
      lo[t[i]].pop();
      ans[tt] = 'A';
    }
  }
  if ((int)(o).size() % 2 == 0) {
    for (int i = 0; i < (int)(th).size(); i++) {
      while (!lo[th[i]].empty()) {
        tt = lo[th[i]].front();
        lo[th[i]].pop();
        ans[tt] = 'A';
      }
    }
  } else {
    tt = lo[th[0]].front();
    lo[th[0]].pop();
    ans[tt] = 'A';
    for (int i = 0; i < (int)(th).size(); i++) {
      while (!lo[th[i]].empty()) {
        tt = lo[th[i]].front();
        lo[th[i]].pop();
        ans[tt] = 'B';
      }
    }
  }
  printf("%s", ans);
}
