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 <bits/stdc++.h>
using namespace std;

signed main () {
    int n, a, b, c;
	cin >> n >> a >> b >> c;
	vector < int > val = {a, b, c};
	vector < string > q(n);
	for (int i = 0; i < n; ++i)
		cin >> q[i];
	string ans;
	for(int i = 0; i < n; ++i) {
		int v1 = q[i][0] - 'A', v2 = q[i][1] - 'A';
		if(val[v1] == 0 && val[v2] == 0) {
			cout << "No" << endl;
			return 0;
		}
		if (val[v1] == 1 && val[v2] == 1) {
			if (i + 1 < n) {
				int v11 = q[i + 1][0] - 'A', v22 = q[i + 1][1] - 'A';
				if (v11 == 3 - v1 - v2 || v22 == 3 - v1 - v2){
					if (v22 == v1 || v11 == v1)
						ans += (char)(v1 + 'A'), val[v1]++, val[v2]--;
					else
						ans += (char)(v2 + 'A'), val[v2]++, val[v1]--;
				}
				else
					ans += (char)(v1 + 'A'), val[v1]++, val[v2]--;
			}
			else
				ans += (char)(v1 + 'A'), val[v1]++, val[v2]--;
		}
		else if (val[v1] > val[v2])
			val[v1]--, val[v2]++, ans += (char)(v2 + 'A');
		else
			val[v2]--, val[v1]++, ans += (char)(v1 + 'A');
	}
	cout << "Yes" << '\n';
	for(auto i: ans)
		cout << i << '\n';
}