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;

template<typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}

signed main(){
  int n,a,b,c;cin>>n>>a>>b>>c;
  vector<string> s(n);
  for(int i=0;i<n;i++)cin>>s[i];
  vector<char> ans(n);
  int f=0;
  int ppa,ppb,ppc,pa=a,pb=b,pc=c;
  for(int i=0;i<n;i++){
    ppa=pa,ppb=pb,ppc=pc;
    pa=a,pb=b,pc=c;
    //cout<<i<<" "<<ppa<<" "<<ppb<<" "<<ppc<<" "<<pa<<" "<<pb<<" "<<pc<<" "<<a<<" "<<b<<" "<<c<<endl;
    if(s[i]=="AB"){
      if((a>b)^(f==2))a--,b++,ans[i]='B';
      else a++,b--,ans[i]='A';
    }
    if(s[i]=="BC"){
      if((b<c)^(f==2))b++,c--,ans[i]='B';
      else b--,c++,ans[i]='C';
    }
    if(s[i]=="AC"){
      if((a<c)^(f==2))a++,c--,ans[i]='A';
      else a--,c++,ans[i]='C';
    }
    if(min({a,b,c})<0){
      if(f||!i)fin("No");
      f=2;i-=2;a=ppa;b=ppb,c=ppc;
    }
    else f/=2;
  }
  cout<<"Yes"<<endl;
  for(int i=0;i<n;i++)cout<<ans[i]<<endl;
}