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;
const int N=1e5+10;
string s[N];
int p[5],ans[N];
int n;
void solve(int i,int x,int y) {
	int x0,y0;
	if (p[x]>p[y]) {
		ans[i]=y; p[x]--; p[y]++;
	}
	else if (p[x]<p[y]) {
		ans[i]=x; p[x]++; p[y]--;
	}
	else if (p[x]>1) {
		ans[i]=x; p[x]++; p[y]--;
	}
	else if (p[x]==0) {
		cout<<"No"<<endl;
		exit(0);
	}
	else {
		if (i==n) {
			ans[i]=x; p[x]++; p[y]--; return;
		}
		x0=s[i+1][0]-'A'; y0=s[i+1][1]-'A';
		if (x0==x&&y0==y) {
			ans[i]=x; p[x]++; p[y]--; return;
		} 
		if (x==x0||x==y0) {
			ans[i]=x; p[x]++; p[y]--; return;
		}
		else {
			ans[i]=y; p[x]--; p[y]++; return;
		}
	}
}
int main() {
	int i;
	ios::sync_with_stdio(false);
	cin>>n>>p[0]>>p[1]>>p[2];
	for (i=1;i<=n;i++) 
		cin>>s[i];
	for (i=1;i<=n;i++) {
		if (s[i]=="AB") solve(i,0,1);
		else if (s[i]=="BC") solve(i,1,2);
		else solve(i,0,2); 
	}
	cout<<"Yes"<<endl;
	for (i=1;i<=n;i++)
		cout<<char(ans[i]+'A')<<endl;
	return 0;
}