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:
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;

int main(){
  int N, a, b, c;
  cin >> N >> a >> b >> c;
  map<char,int> M;
  M['A'] = a;
  M['B'] = b;
  M['C'] = c;
  vector<string> S(N);
  for(int i = 0; i < N; ++i){
    cin >> S[i];
  }
  vector<char> ans;
  for(int i = 0; i < N; ++i){
    string s = S[i];
    char l = s[0], r = s[1];
    if(M[l] == 0 and M[r] == 0){
      puts("No");
      return 0;
    }
    if(M[l] == 1 and M[r] == 1){
      if(i+1 < N){
        if(count(S[i+1].begin(), S[i+1].end(),l)) ++M[l],--M[r],ans.push_back(l);
        else if(count(S[i+1].begin(), S[i+1].end(),r)) ++M[r],--M[l],ans.push_back(r);
        else ++M[l],--M[r],ans.push_back(l);
      }else
        ++M[l],--M[r],ans.push_back(l);
    }else{
      if(M[l] >= M[r]){
        ++M[r],--M[l],ans.push_back(r);
      }else{
        ++M[l],--M[r],ans.push_back(l);
      }
    }
  }
  puts("Yes");
  for(auto e : ans)
    cout << e << "\n";
}