#include <iostream>
#include <cstdio>
#define N 100005
using namespace std;

int n, a[3];
char s[N][5], ans[N];

void f(int p, int q, int r) {
	ans[p] = q + 'A';
	a[q]++, a[r]--;
}

int main() {
	int i, x, y, t;
	cin >> n >> a[0] >> a[1] >> a[2];
	for(i=0; i<n; i++) scanf("%s", &s[i]);
	for(i=0; i<n; i++) {
		x = s[i][0] - 'A';
		y = s[i][1] - 'A';
		if(!a[x] && !a[y]) return puts("No"), 0;
		if(!a[x]) f(i, x, y);
		else if(!a[y]) f(i, y, x);
		else if(a[x]==1 && a[y]==1) {
			if(x==0) {
				t = s[i+1][0] - 'A';
				t==0 ? f(i, x, y) : f(i, y, x);
			} else {
				t = s[i+1][1] - 'A';
				t==2 ? f(i, y, x) : f(i, x, y);
			}
		} else a[x]<a[y] ? f(i, x, y) : f(i, y, x);
	}
	puts("Yes");
	for(i=0; i<n; i++) printf("%c\n", ans[i]);
	return 0;
}
### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include <iostream>
#include <cstdio>
#define N 100005
using namespace std;

int n, a[3];
char s[N][5], ans[N];

void f(int p, int q, int r) {
	ans[p] = q + 'A';
	a[q]++, a[r]--;
}

int main() {
	int i, x, y, t;
	cin >> n >> a[0] >> a[1] >> a[2];
	for(i=0; i<n; i++) scanf("%s", &s[i]);
	for(i=0; i<n; i++) {
		x = s[i][0] - 'A';
		y = s[i][1] - 'A';
		if(!a[x] && !a[y]) return puts("No"), 0;
		if(!a[x]) f(i, x, y);
		else if(!a[y]) f(i, y, x);
		else if(a[x]==1 && a[y]==1) {
			if(x==0) {
				t = s[i+1][0] - 'A';
				t==0 ? f(i, x, y) : f(i, y, x);
			} else {
				t = s[i+1][1] - 'A';
				t==2 ? f(i, y, x) : f(i, x, y);
			}
		} else a[x]<a[y] ? f(i, x, y) : f(i, y, x);
	}
	puts("Yes");
	for(i=0; i<n; i++) printf("%c\n", ans[i]);
	return 0;
}
```