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;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;

	int a, b, c;
	cin >> a >> b >> c;

	vector<string> s(n);
	for(int i = 0; i < n; ++i) {
		cin >> s[i];
	}

	int A = 0;
	int B = 0;
	int C = 0;

	vector<char> ans(n, 'D');

	bool ok = true;

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

	dfs(0, a, b, c);

	cout << "No" << '\n';

	return 0;
}