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 = 105;
const int INF = 0x3f3f3f3f;
int cnt[maxn], a[maxn], vis[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  int x = 0, y = 0;
  for (int i = 1; i <= n; i++) {
    if (cnt[a[i]] == 1) {
      if (x < y) {
        vis[i] = 0;
        x++;
      } else {
        vis[i] = 1;
        y++;
      }
    }
  }
  if (x < y) {
    for (int i = 1; i <= n; i++) {
      if (cnt[a[i]] > 2) {
        x++;
        vis[i] = 0;
        for (int j = i + 1; j <= n; j++)
          if (a[j] == a[i]) vis[j] = 1;
        break;
      }
    }
  } else if (x > y) {
    for (int i = 1; i <= n; i++) {
      if (cnt[a[i]] > 2) {
        y++;
        vis[i] = 1;
        for (int j = i + 1; j <= n; j++)
          if (a[j] == a[i]) vis[j] = 0;
        break;
      }
    }
  }
  if (x != y)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++)
      if (!vis[i])
        cout << "A";
      else
        cout << "B";
    cout << endl;
  }
}