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 <cstdio>
#include <string>
#include <vector>
#include <array>
using namespace std;

int main() {

	int N;
	array<int, 3> xs;
	vector<pair<int,int>> queries;
	vector<int> ans;
	scanf("%d%d%d%d", &N, &xs[0], &xs[1], &xs[2]);
	queries.resize(N);

	for(int i = 0; i < N; i++) {
		char buf[16];
		scanf("%s", buf);
		int i1 = buf[0] - 'A';
		int i2 = buf[1] - 'A';
		queries[i] = {i1, i2};
	}

	for(int step = 0; step < N; step++) {

		int i1 = queries[step].first;
		int i2 = queries[step].second;

		if(xs[i1] == 0 && xs[i2] == 0) {
			printf("No\n");
			return 0;
		}

		if(xs[i1] == 1 && xs[i2] == 1 && step < N - 1) {

			int j1 = queries[step + 1].first;
			int j2 = queries[step + 1].second;

			if(i1 == j1 || i1 == j2) {
				swap(i1, i2);
			}

		} else {
			if(xs[i1] < xs[i2]) {
				swap(i1, i2);
			}
		}

		xs[i1] -= 1;
		xs[i2] += 1;
		ans.push_back(i2);
	}

	printf("Yes\n");
	for(int i = 0; i < N; i++) {
		printf("%c\n", ans[i] + 'A');
	}

	return 0;
}