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;
#define modulo N
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000005



int main(){
	
	int N;
	cin>>N;
	
	vector<int> n(3);
	cin>>n[0]>>n[1]>>n[2];
	
	vector<string> S(N);
	for(int i=0;i<N;i++)cin>>S[i];
	
	string ans = "";
	
	for(int i=0;i<N;i++){
		pair<int,int> p;
		if(S[i]=="AB")p={0,1};
		else if(S[i]=="BC")p={1,2};
		else p={0,2};
		
		if(n[p.first]==0&&n[p.second]==0){
			cout<<"No"<<endl;
			return 0;
		}
		
		if(n[p.first]==1&&n[p.second]==1){
			if(i==N-1||(S[i+1][0]!=S[i][0]&&S[i+1][1]!=S[i][0])){
				n[p.first]--;
				n[p.second]++;
				ans+=S[i][1];
			}
			else{
				n[p.first]++;
				n[p.second]--;
				ans+=S[i][0];
			}
		}
		else{
			if(n[p.first]>=n[p.second]){
				ans+=S[i][1];
				n[p.first]--;
				n[p.second]++;
			}
			else{
				ans+=S[i][0];
				n[p.first]++;
				n[p.second]--;
			}
		}
	}
	
	cout<<"Yes"<<endl;
	for(int i=0;i<N;i++)cout<<ans[i]<<endl;
	
	
	return 0;
}