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;
int n,var[4],chs[100100],bck[100100][4];
string str[100100];
void bao(int id){
	if(id==n+1){
		cout<<"Yes"<<endl;
		for(int i=1;i<=n;i++) cout<<(char)(chs[i]+'A'-1)<<endl;
		exit(0);
	}
	int x=str[id][0]-'A'+1;
	int y=str[id][1]-'A'+1;
	if(var[x]>=bck[id][x]){
		var[x]--;
		var[y]++;
		chs[id]=y;
		bao(id+1);
		var[y]--;
		var[x]++;
		return;
	}
	if(var[y]>=bck[id][y]){
		var[y]--;
		var[x]++;
		chs[id]=x;
		bao(id+1);
		var[x]--;
		var[y]++;
		return;
	}
	if(var[x]){
		var[x]--;
		var[y]++;
		chs[id]=y;
		bao(id+1);
		var[y]--;
		var[x]++;
	}
	if(var[y]){
		var[y]--;
		var[x]++;
		chs[id]=x;
		bao(id+1);
		var[x]--;
		var[y]++;
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>var[1]>>var[2]>>var[3];
	for(int i=1;i<=n;i++){
		cin>>str[i];
	}
	int xx,yy;
	for(int i=n;i>=1;i--){
		xx=str[i][0]-'A'+1;
		yy=str[i][1]-'A'+1;
		bck[i][1]=bck[i+1][1];
		bck[i][2]=bck[i+1][2];
		bck[i][3]=bck[i+1][3];
		bck[i][xx]++;
		bck[i][yy]++;
	}
	bao(1);
	cout<<"No"<<endl;
	return 0;
}