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>
using namespace std;
#define OPTIMISE ios::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);

int a,b,c,n;
const int M = 1e5+100;
string s[M];
char ans[M];

void dfs(int i, int a, int b, int c) {
	if (i == n) {
		cout << "Yes\n";
		for (int i = 0; i < n; i++)
			cout << ans[i] << "\n";
		exit(0);
	}
	if (s[i] == "AB") {
		if (a) ans[i] = 'B', dfs(i+1,a-1,b+1,c);
		if (b) ans[i] = 'A', dfs(i+1,a+1,b-1,c);
	}
	if (s[i] == "BC") {
		if (b) ans[i] = 'C', dfs(i+1,a,b-1,c+1);
		if (c) ans[i] = 'B', dfs(i+1,a,b+1,c-1);
	}
	if (s[i] == "AC") {
		if (a) ans[i] = 'C', dfs(i+1,a-1,b,c+1);
		if (c) ans[i] = 'A', dfs(i+1,a+1,b,c-1);
	}
}

int main() {
	OPTIMISE;

	cin >> n >> a >> b >> c;
	for (int i = 0; i < n; i++)
		cin >> s[i];
	dfs(0,a,b,c);
	cout << "No";
}