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 N = 1e5 + 11, mod = 1e9 + 7;
const int MAX = 1e9 + 11;
const int INF1 = 2e9 + 11;
const long long INF2 = 2e18 + 11;
const int base = 500;
const int P = 31;
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
const double EPS = 1e-12;
const double PI = acos(-1.0);
int n, s[111], ct[111];
int main() {
  if (fopen("peacefulsets"
            ".in",
            "r"))
    freopen(
        "peacefulsets"
        ".in",
        "r", stdin),
        freopen(
            "peacefulsets"
            ".out",
            "w", stdout);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  set<int> r;
  for (int i = 1; i <= n; i += 1) cin >> s[i], ++ct[s[i]], r.insert(s[i]);
  set<int> st;
  int cnt = 0;
  for (auto& it : r)
    if (ct[it] == 1)
      st.insert(it);
    else if (ct[it] != 2)
      ++cnt;
  if (st.size() % 2 && !cnt) return !(cout << "NO");
  cout << "YES\n";
  bool t = 0;
  bool x = 1;
  for (int i = 1; i <= n; i += 1) {
    if (!t && st.size() % 2 && ct[s[i]] > 2) {
      t = 1;
      cout << 'B';
    } else {
      if (ct[s[i]] == 1) {
        cout << (x ? 'A' : 'B');
        x ^= 1;
      } else {
        cout << 'A';
      }
    }
  }
  return 0;
}