#include <bits/stdc++.h>
using namespace std;
using ll=long long;
int main(){
  map<char,ll> M;
  ll N; cin>>N>>M['A']>>M['B']>>M['C'];
  if(M['A']+M['B']+M['C']==0){
    cout<<"No"<<endl; return 0;
  }
  vector<string> Z(N+1,"xx");
  vector<char> ans(0,0);
  bool out=false;
  for(int i=0; i<N; i++) cin>>Z[i];
  for(int i=0; i<N; i++){
    char x=Z[i][0], y=Z[i][1];
    char xx=Z[i+1][0], yy=Z[i+1][1];
    if(M[x]==0 && M[y]==0){
      out=true;
      break;
    }
    else if(M[x]==0) M[x]++, M[y]--, ans.emplace_back(x);
    else if(M[y]==0) M[x]--, M[y]++, ans.emplace_back(y);
    else if(M[x]==1 && M[y]==1){
      if(xx=='x') ans.emplace_back(x);//おしまい
      else if(x==xx || x==yy) M[x]++, M[y]--, ans.emplace_back(x);
      else if(y==xx || y==yy) M[x]--, M[y]++, ans.emplace_back(y);
    }
    else{
      if(M[x]>=M[y]) M[x]--,M[y]++,ans.emplace_back(y);
      else M[x]++,M[y]--,ans.emplace_back(x);
    }
  }
  ll asz=ans.size();
  if(out) cout<<"No"<<endl;
  else{
    cout<<"Yes"<<endl;
    for(int i=0; i<asz; i++) cout<<ans[i]<<endl;
  }
}

### Prompt

Please create a solution in Cpp to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
int main(){
  map<char,ll> M;
  ll N; cin>>N>>M['A']>>M['B']>>M['C'];
  if(M['A']+M['B']+M['C']==0){
    cout<<"No"<<endl; return 0;
  }
  vector<string> Z(N+1,"xx");
  vector<char> ans(0,0);
  bool out=false;
  for(int i=0; i<N; i++) cin>>Z[i];
  for(int i=0; i<N; i++){
    char x=Z[i][0], y=Z[i][1];
    char xx=Z[i+1][0], yy=Z[i+1][1];
    if(M[x]==0 && M[y]==0){
      out=true;
      break;
    }
    else if(M[x]==0) M[x]++, M[y]--, ans.emplace_back(x);
    else if(M[y]==0) M[x]--, M[y]++, ans.emplace_back(y);
    else if(M[x]==1 && M[y]==1){
      if(xx=='x') ans.emplace_back(x);//おしまい
      else if(x==xx || x==yy) M[x]++, M[y]--, ans.emplace_back(x);
      else if(y==xx || y==yy) M[x]--, M[y]++, ans.emplace_back(y);
    }
    else{
      if(M[x]>=M[y]) M[x]--,M[y]++,ans.emplace_back(y);
      else M[x]++,M[y]--,ans.emplace_back(x);
    }
  }
  ll asz=ans.size();
  if(out) cout<<"No"<<endl;
  else{
    cout<<"Yes"<<endl;
    for(int i=0; i<asz; i++) cout<<ans[i]<<endl;
  }
}

```