Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*input
8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB
*/

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

using namespace std;

void run() {
	int n, arr[3], i;
	cin >> n >> arr[0] >> arr[1] >> arr[2];
	vector<string> ops(n);
	vector<int> ret(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) {
			arr[x]++;
			arr[y]--;
			ret[i] = x;
		}
		else if(arr[y] == 0) {
			arr[x]--;
			arr[y]++;
			ret[i] = y;
		}
		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) {
					arr[x]++;
					ret[i] = x;
				}
				else {
					arr[y]++;
					ret[i] = y;
				}
			}
			else {
				ret[i] = x;
			}
		}
		else {
			if(arr[x] < arr[y]) {
				arr[x]++;
				ret[i] = x;
			}
			else {
				arr[y]++;
				ret[i] = y;
			}
		}
	}
	cout << "Yes\n";
	for(int x : ret)
		cout << (char) (x + 'A') << "\n";
}

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