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>
#define st first
#define nd second

using ll = long long ;
using namespace std;
using pii = pair<int,int>;
const int N = 1e5 + 10;
char s[N][5];
int ans[N];
int main(){
	int n, a[3];
	scanf("%d %d %d %d", &n, &a[0], &a[1], &a[2]);
	bool io = (1LL * a[0] + a[1] + a[2]) >= 3LL;
	for(int i = 1; i <= n; i++){
		scanf("%s", s[i]);
	}

	for(int i = 1; i <= n; i++){
		int pt1 = s[i][0] - 'A', pt2 = s[i][1] - 'A';
		if(a[pt1] == 0 && a[pt2] == 0) return !printf("No");
		if(io || (i == n || a[pt1] != 1 || a[pt2] != 1)){
			if(a[pt1] > a[pt2]) a[pt2]++, a[pt1]--, ans[i] = pt2;
			else a[pt1]++, a[pt2]--, ans[i] = pt1;
		}
		else{
			if(s[i + 1][0] == s[i][0] && s[i + 1][1] == s[i][1]) a[pt2]++, a[pt1]--, ans[i] = pt2;
			else if(s[i + 1][0] == s[i][0] || s[i + 1][1] == s[i][0]) a[pt1]++, a[pt2]--, ans[i] = pt1;
			else a[pt2]++, a[pt1]--, ans[i] = pt2;
		}
	}
	printf("Yes\n");
	for(int i = 1; i <= n; i++){ 
		printf("%c\n", 'A' + ans[i]);
	}

}