#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using vll = vector<ll> ;
#define all(v) v.begin(),v.end()

int main(){
  ll N,A,B,C;cin>>N>>A>>B>>C;
  vector<string> s(N+1);rep(i,N)cin>>s[i];
  char ans[N];
  bool flag=true;
  rep(i,N){
    if(A==0 && B==0 && C==0){flag=false;break;}
    if(s[i]=="AB"){
      if(A==0 && B==0){flag=false;break;}
      else if(A>B || A==B&&s[i+1]=="BC"){ans[i]='B';++B;--A;}
      else {ans[i]='A';++A;--B;}
    }
    if(s[i]=="BC"){
      if(C==0 && B==0){flag=false;break;}
      else if(B>C || B==C&&s[i+1]=="AC"){ans[i]='C';++C;--B;}
      else {ans[i]='B';++B;--C;}
    }
    if(s[i]=="AC"){
      if(A==0 && C==0){flag=false;break;}
      else if(C>A || C==A&&s[i+1]=="AB" && B==0){ans[i]='A';++A;--C;}
      else {ans[i]='C';++C;--A;}      
    }
    //cout << "A" << A << "B" << B << "C" << C << endl;
  }
  if(flag){cout<<"Yes"<<endl;rep(i,N)cout<<ans[i]<<endl;}
  else cout<<"No"<<endl;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using vll = vector<ll> ;
#define all(v) v.begin(),v.end()

int main(){
  ll N,A,B,C;cin>>N>>A>>B>>C;
  vector<string> s(N+1);rep(i,N)cin>>s[i];
  char ans[N];
  bool flag=true;
  rep(i,N){
    if(A==0 && B==0 && C==0){flag=false;break;}
    if(s[i]=="AB"){
      if(A==0 && B==0){flag=false;break;}
      else if(A>B || A==B&&s[i+1]=="BC"){ans[i]='B';++B;--A;}
      else {ans[i]='A';++A;--B;}
    }
    if(s[i]=="BC"){
      if(C==0 && B==0){flag=false;break;}
      else if(B>C || B==C&&s[i+1]=="AC"){ans[i]='C';++C;--B;}
      else {ans[i]='B';++B;--C;}
    }
    if(s[i]=="AC"){
      if(A==0 && C==0){flag=false;break;}
      else if(C>A || C==A&&s[i+1]=="AB" && B==0){ans[i]='A';++A;--C;}
      else {ans[i]='C';++C;--A;}      
    }
    //cout << "A" << A << "B" << B << "C" << C << endl;
  }
  if(flag){cout<<"Yes"<<endl;rep(i,N)cout<<ans[i]<<endl;}
  else cout<<"No"<<endl;
}
```