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 <iostream>
#include <string>
#include <vector>
using namespace std;

int main() {
	int N;
	cin >> N;
	vector<int> ABC(3);
	for (int& i : ABC) cin >> i;
	vector<string> s(N);
	for (string& i : s) cin >> i;
	bool a = true;
	string b;
	for (vector<string>::iterator i = s.begin(); i != s.end(); i++) {
		if (!ABC[i->front() - 'A']) {
			if (!ABC[i->back() - 'A']) {
				a = false;
				break;
			}
			ABC[i->front() - 'A']++;
			ABC[i->back() - 'A']--;
			b += i->front();
		}
		else if (!ABC[i->back() - 'A']) {
			ABC[i->front() - 'A']--;
			ABC[i->back() - 'A']++;
			b += i->back();
		}
		else {
			if (i == s.end() - 1) {
				ABC[i->front() - 'A']++;
				ABC[i->back() - 'A']--;
				b += i->front();
			}
			else {
				if (i->front() == (i + 1)->front() || i->front() == (i + 1)->back()) {
					ABC[i->front() - 'A']++;
					ABC[i->back() - 'A']--;
					b += i->front();
				}
				else {
					ABC[i->front() - 'A']--;
					ABC[i->back() - 'A']++;
					b += i->back();
				}
			}
		}
	}
	if (a) {
		cout << "Yes" << endl;
		for (char& i : b) cout << i << endl;
	}
	else cout << "No";
}