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;
int n, f[102], a[103], ans[104];
set<int> s1, s2;
vector<int> v[102];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    f[a[i]]++;
    if (f[a[i]] > 2) {
      s2.insert(a[i]);
    }
    v[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    if (f[a[i]] == 1) {
      s1.insert(a[i]);
    }
  }
  if ((s1.size()) % 2 && !s2.size()) {
    return puts("NO"), 0;
  } else if (s1.size() % 2) {
    int cnt = (s1.size()) / 2;
    auto it = s1.begin();
    int x = 0;
    while (x < cnt) {
      x++;
      int pos = v[*it][0];
      ans[pos] = 1;
      it = s1.erase(it);
    }
    int z = *(s2.begin());
    ans[v[z].back()] = 1;
  } else {
    int cnt = (s1.size()) / 2;
    auto it = s1.begin();
    int x = 0;
    while (x < cnt) {
      x++;
      int pos = v[*it][0];
      ans[pos] = 1;
      it = s1.erase(it);
    }
  }
  puts("YES");
  for (int i = 0; i < n; i++) {
    if (ans[i])
      cout << "A";
    else
      cout << "B";
  }
}
