Problem: There is a game that involves three variables, denoted A, B, and C.

As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other.

After each choice, none of A, B, and C should be negative.

Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Constraints

* 1 \leq N \leq 10^5
* 0 \leq A,B,C \leq 10^9
* N, A, B, C are integers.
* s_i is `AB`, `AC`, or `BC`.

Input

Input is given from Standard Input in the following format:


N A B C
s_1
s_2
:
s_N


Output

If it is possible to make N choices under the condition, print `Yes`; otherwise, print `No`.

Also, in the former case, show one such way to make the choices in the subsequent N lines. The (i+1)-th line should contain the name of the variable (`A`, `B`, or `C`) to which you add 1 in the i-th choice.

Examples

Input

2 1 3 0
AB
AC


Output

Yes
A
C


Input

3 1 0 0
AB
BC
AB


Output

No


Input

1 0 9 0
AC


Output

No


Input

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB


Output

Yes
C
B
B
C
C
B
A
A

Solution (in CPP):

Solution:
/**
 *    author:  tourist
 *    created: 11.05.2020 13:47:54       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, a, b, c;
  cin >> n >> a >> b >> c;
  vector<string> s(n);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  string res(n, ' ');
  for (int i = 0; i < n; i++) {
    int& x = (s[i][0] == 'A' ? a : (s[i][0] == 'B' ? b : c));
    int& y = (s[i][1] == 'A' ? a : (s[i][1] == 'B' ? b : c));
    if (x + y == 0) {
      cout << "No" << '\n';
      return 0;
    }
    int id = -1;
    if (x != 1 || y != 1) {
      id = (x > y ? 1 : 0);
    } else {
      id = (i == n - 1 ? 0 : (s[i + 1].find(s[i][0]) != string::npos ? 0 : 1));
    }
    res[i] = s[i][id];
    x += (id == 0 ? 1 : -1);
    y += (id == 1 ? 1 : -1);
  }
  cout << "Yes" << '\n';
  for (char ch : res) {
    cout << ch << '\n';
  }
  return 0;
}