#include<bits/stdc++.h>
using namespace std;
int main() {
  int64_t n,a,b,c,i;
  cin>>n>>a>>b>>c;
  vector<string> s(n),d(n);
  for(i=0;i<n;i++) cin>>s.at(i);
  for(i=0;i<n;i++){
    if(s.at(i)=="AB"){
      if(a==0 && b==0) break;
      else if(a<b) {a++; b--; d.at(i)="A";}
      else if(b<a) {a--; b++; d.at(i)="B";}
      else if(a==b){
        if(i<n-1 && s.at(i+1)=="BC") {a--; b++; d.at(i)="B";}
        else {a++; b--; d.at(i)="A";}
      }
    }
    else if(s.at(i)=="AC"){
      if(a==0 && c==0) break;
      else if(a<c) {a++; c--; d.at(i)="A";}
      else if(c<a) {a--; c++; d.at(i)="C";}
      else if(a==c){
        if(i<n-1 && s.at(i+1)=="BC") {a--; c++; d.at(i)="C";}
        else {a++; c--; d.at(i)="A";}
      }
    }
    else if(s.at(i)=="BC"){
      if(b==0 && c==0) break;
      else if(b<c) {b++; c--; d.at(i)="B";}
      else if(c<b) {b--; c++; d.at(i)="C";}
      else if(b==c){
        if(i<n-1 && s.at(i+1)=="AC") {b--; c++; d.at(i)="C";}
        else {b++; c--; d.at(i)="B";}
      }
    }
  }
  if (i<n) cout<<"No"<<endl;
  else {
    cout<<"Yes"<<endl;
    for(i=0;i<n;i++) cout<<d.at(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;
int main() {
  int64_t n,a,b,c,i;
  cin>>n>>a>>b>>c;
  vector<string> s(n),d(n);
  for(i=0;i<n;i++) cin>>s.at(i);
  for(i=0;i<n;i++){
    if(s.at(i)=="AB"){
      if(a==0 && b==0) break;
      else if(a<b) {a++; b--; d.at(i)="A";}
      else if(b<a) {a--; b++; d.at(i)="B";}
      else if(a==b){
        if(i<n-1 && s.at(i+1)=="BC") {a--; b++; d.at(i)="B";}
        else {a++; b--; d.at(i)="A";}
      }
    }
    else if(s.at(i)=="AC"){
      if(a==0 && c==0) break;
      else if(a<c) {a++; c--; d.at(i)="A";}
      else if(c<a) {a--; c++; d.at(i)="C";}
      else if(a==c){
        if(i<n-1 && s.at(i+1)=="BC") {a--; c++; d.at(i)="C";}
        else {a++; c--; d.at(i)="A";}
      }
    }
    else if(s.at(i)=="BC"){
      if(b==0 && c==0) break;
      else if(b<c) {b++; c--; d.at(i)="B";}
      else if(c<b) {b--; c++; d.at(i)="C";}
      else if(b==c){
        if(i<n-1 && s.at(i+1)=="AC") {b--; c++; d.at(i)="C";}
        else {b++; c--; d.at(i)="B";}
      }
    }
  }
  if (i<n) cout<<"No"<<endl;
  else {
    cout<<"Yes"<<endl;
    for(i=0;i<n;i++) cout<<d.at(i)<<endl;
  }
}
```