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

#define rep(i, n) for(int i=0; i<n; ++i)

int main(){
   int n, a[3];
   cin>>n>>a[0]>>a[1]>>a[2];
   string s[100000];
   rep(i, n) cin>>s[i];

   int ans[100000];
   rep(i, n){
      int e, f;
      e=s[i][0]-'A';
      f=s[i][1]-'A';
      if(a[e]==0 && a[f]==0){
         cout<<"No"<<endl;
         return 0;
      }
      else if(a[e]==0){
         ans[i]=e; a[e]++; a[f]--;
      }
      else if(a[f]==0){
         ans[i]=f; a[f]++; a[e]--;
      }
      else if(a[e]==1 && a[f]==1 && i!=n-1){
         int g, h;
         g=s[i+1][0]-'A';
         h=s[i+1][1]-'A';
         if(e==g || e==h){
            ans[i]=e; a[e]++; a[f]--;
         }
         else{
            ans[i]=f; a[f]++; a[e]--;
         }
      }
      else{
         if(a[e]<a[f]){
            ans[i]=e; a[e]++; a[f]--;
         }
         else{
            ans[i]=f; a[f]++; a[e]--;
         }
      }
   }

   cout << "Yes" << endl;
   rep(i, n) cout << (char)('A'+ans[i]) << endl;
   return 0;
}

### 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;

#define rep(i, n) for(int i=0; i<n; ++i)

int main(){
   int n, a[3];
   cin>>n>>a[0]>>a[1]>>a[2];
   string s[100000];
   rep(i, n) cin>>s[i];

   int ans[100000];
   rep(i, n){
      int e, f;
      e=s[i][0]-'A';
      f=s[i][1]-'A';
      if(a[e]==0 && a[f]==0){
         cout<<"No"<<endl;
         return 0;
      }
      else if(a[e]==0){
         ans[i]=e; a[e]++; a[f]--;
      }
      else if(a[f]==0){
         ans[i]=f; a[f]++; a[e]--;
      }
      else if(a[e]==1 && a[f]==1 && i!=n-1){
         int g, h;
         g=s[i+1][0]-'A';
         h=s[i+1][1]-'A';
         if(e==g || e==h){
            ans[i]=e; a[e]++; a[f]--;
         }
         else{
            ans[i]=f; a[f]++; a[e]--;
         }
      }
      else{
         if(a[e]<a[f]){
            ans[i]=e; a[e]++; a[f]--;
         }
         else{
            ans[i]=f; a[f]++; a[e]--;
         }
      }
   }

   cout << "Yes" << endl;
   rep(i, n) cout << (char)('A'+ans[i]) << endl;
   return 0;
}

```