// In the name of God

#include <bits/stdc++.h>

using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	vector<int> pt(3);
	int n;
	cin >> n >> pt[0] >> pt[1] >> pt[2];
	vector<pair<int,int>> q;
	for(int i = 0; i < n; i++) {
		string s;
		cin >> s;
		if(s == "AB") q.push_back(make_pair(0, 1));
		if(s == "AC") q.push_back(make_pair(0, 2));
		if(s == "BC") q.push_back(make_pair(1, 2));
	}
	vector<int> ans;
	for(int i = 0; i < n; i++) {
		int p1 = q[i].first, p2 = q[i].second;
		if(pt[p1] == 0 && pt[p2] == 0) {
			cout << "No";
			exit(0);
		}
		if(pt[p1] == pt[p2]) {
			if(i != n-1 && (p1 == q[i+1].first || p1 == q[i+1].second)) {
				pt[p1]++;
				pt[p2]--;
				ans.push_back(p1);
			}
			else {
				pt[p2]++;
				pt[p1]--;
				ans.push_back(p2);
			}
		}
		else {
			if(pt[p1] > pt[p2]) {
				pt[p2]++;
				pt[p1]--;
				ans.push_back(p2);
			}
			else {
				pt[p1]++;
				pt[p2]--;
				ans.push_back(p1);
			}
		}
	}
	cout << "Yes\n";
	for(int x : ans) {
		if(x == 0) cout << "A\n";
		if(x == 1) cout << "B\n";
		if(x == 2) cout << "C\n";
	}
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following 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

### Response

```cpp
// In the name of God

#include <bits/stdc++.h>

using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	vector<int> pt(3);
	int n;
	cin >> n >> pt[0] >> pt[1] >> pt[2];
	vector<pair<int,int>> q;
	for(int i = 0; i < n; i++) {
		string s;
		cin >> s;
		if(s == "AB") q.push_back(make_pair(0, 1));
		if(s == "AC") q.push_back(make_pair(0, 2));
		if(s == "BC") q.push_back(make_pair(1, 2));
	}
	vector<int> ans;
	for(int i = 0; i < n; i++) {
		int p1 = q[i].first, p2 = q[i].second;
		if(pt[p1] == 0 && pt[p2] == 0) {
			cout << "No";
			exit(0);
		}
		if(pt[p1] == pt[p2]) {
			if(i != n-1 && (p1 == q[i+1].first || p1 == q[i+1].second)) {
				pt[p1]++;
				pt[p2]--;
				ans.push_back(p1);
			}
			else {
				pt[p2]++;
				pt[p1]--;
				ans.push_back(p2);
			}
		}
		else {
			if(pt[p1] > pt[p2]) {
				pt[p2]++;
				pt[p1]--;
				ans.push_back(p2);
			}
			else {
				pt[p1]++;
				pt[p2]--;
				ans.push_back(p1);
			}
		}
	}
	cout << "Yes\n";
	for(int x : ans) {
		if(x == 0) cout << "A\n";
		if(x == 1) cout << "B\n";
		if(x == 2) cout << "C\n";
	}
	return 0;
}

```