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 <bits/stdc++.h>
using namespace std;

vector <int64_t>num(3);
vector<int>ans(1000010);

void jadge(int i,int p,int q){
  ans.at(i)=p;
  num.at(p)++;
  num.at(q)--;
}

int main()
{
  int N,p,q;
  cin >>N>>num.at(0)>>num.at(1)>>num.at(2);
 
  vector<vector<int>>x(N,vector<int>(2));
  for(int i=0;i<N;i++){
    string s;
    cin >> s;
    x.at(i).at(0)=s.at(0)-'A';
    x.at(i).at(1)=s.at(1)-'A';
  }
  for(int i=0;i<N;i++){
    p=x.at(i).at(0);
    q=x.at(i).at(1);
    if(num.at(p)==0&&num.at(q)==0){
      cout <<"No"<<endl;
      return 0;
    }
    else if(num.at(p)==0)
      jadge(i,p,q);
    else if(num.at(q)==0)
      jadge(i,q,p);
    else if(i!=N-1&&p==3-x.at(i+1).at(0)-x.at(i+1).at(1))jadge(i,q,p);
    else jadge(i,p,q);
  }
  cout <<"Yes"<<endl;
  for(int i=0;i<N;i++){
    cout <<(char)(ans.at(i)+'A')<<endl;
  }
  return 0;
}