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;
vector<vector<int>> g;
vector<int> ch;
vector<bool> vis1;
vector<int> par(1000000);
vector<int> sz(1000000);
void print(vector<int> v) {
  vector<int>::iterator it = v.begin();
  while (it != v.end()) {
    cout << *it << " ";
    it++;
  }
  cout << endl;
}
void create(int n) {
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
int fp(int n) {
  if (par[n] == n) return n;
  return par[n] = fp(par[n]);
}
void merg(int x, int y) {
  x = fp(x);
  y = fp(y);
  if (sz[x] < sz[y]) {
    swap(x, y);
  }
  par[y] = x;
  sz[x] += sz[y];
}
int main() {
  int n;
  cin >> n;
  vector<int> occ(101);
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    occ[v[i]]++;
  }
  int nc = 0;
  int gtt = -1;
  map<int, int> M;
  for (int i = 1; i <= 100; i++) {
    if (occ[i] == 1) {
      nc++;
      M[i] = 1;
    } else if (occ[i] > 2) {
      gtt = i;
    }
  }
  if (nc % 2 == 0) {
    string s;
    int prev = 66;
    for (int i = 0; i < n; i++) {
      if (M[v[i]] == 1) {
        s += char(131 - prev);
        prev = 131 - prev;
      } else {
        s += 'A';
      }
    }
    cout << "YES" << endl << s << endl;
  } else if (nc % 2 == 1 && gtt != -1) {
    int x;
    bool d = 0;
    string s;
    int prev = 66;
    for (int i = 0; i < n; i++) {
      if (M[v[i]] == 1 || (v[i] == gtt && d == 0)) {
        s += char(131 - prev);
        prev = 131 - prev;
        if (v[i] == gtt) {
          d = 1;
          x = 131 - prev;
        }
      } else if (v[i] == gtt && d == 1) {
        s += char(x);
      } else {
        s += 'A';
      }
    }
    cout << "YES" << endl << s << endl;
  } else {
    cout << "NO" << endl;
  }
}