/*input
8 6 1 1
AC
BC
AB
BC
AC
BC
AB
AB
*/

#include <bits/stdc++.h>
#define ll long long

using namespace std;

vector<int> ret, arr(3);

void update(int x, int y, int index) {
	ret[index] = x;
	arr[x]++;
	arr[y]--;
}

void run() {
	int n, i;
	cin >> n >> arr[0] >> arr[1] >> arr[2];
	vector<string> ops(n);
	ret.resize(n);
	for(auto &s : ops)
		cin >> s;
	for(i=0; i<n; i++) {
		int x = ops[i][0] - 'A', y = ops[i][1] - 'A';
		if(arr[x] <= 0  &&  arr[y] <= 0) {
			cout << "No\n";
			return;
		}
		if(arr[x] == 0)
			update(x, y, i);
		else if(arr[y] == 0)
			update(y, x, i);
		else if(arr[x] == 1  &&  arr[y] == 1) {
			if(i + 1 < n) {
				if(((ops[i + 1][0]-'A') == x)  ||  ((ops[i + 1][1]-'A') == x))
					update(x, y, i);
				else
					update(y, x, i);
			}
			else
				update(x, y, i);
		}
		else {
			if(arr[x] <= arr[y])
				update(x, y, i);
			else
				update(y, x, i);
		}
	}
	cout << "Yes\n";
	for(int x : ret)
		cout << (char) (x + 'A') << "\n";
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);		
	run();
}

### 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
/*input
8 6 1 1
AC
BC
AB
BC
AC
BC
AB
AB
*/

#include <bits/stdc++.h>
#define ll long long

using namespace std;

vector<int> ret, arr(3);

void update(int x, int y, int index) {
	ret[index] = x;
	arr[x]++;
	arr[y]--;
}

void run() {
	int n, i;
	cin >> n >> arr[0] >> arr[1] >> arr[2];
	vector<string> ops(n);
	ret.resize(n);
	for(auto &s : ops)
		cin >> s;
	for(i=0; i<n; i++) {
		int x = ops[i][0] - 'A', y = ops[i][1] - 'A';
		if(arr[x] <= 0  &&  arr[y] <= 0) {
			cout << "No\n";
			return;
		}
		if(arr[x] == 0)
			update(x, y, i);
		else if(arr[y] == 0)
			update(y, x, i);
		else if(arr[x] == 1  &&  arr[y] == 1) {
			if(i + 1 < n) {
				if(((ops[i + 1][0]-'A') == x)  ||  ((ops[i + 1][1]-'A') == x))
					update(x, y, i);
				else
					update(y, x, i);
			}
			else
				update(x, y, i);
		}
		else {
			if(arr[x] <= arr[y])
				update(x, y, i);
			else
				update(y, x, i);
		}
	}
	cout << "Yes\n";
	for(int x : ret)
		cout << (char) (x + 'A') << "\n";
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);		
	run();
}

```