#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

void No() {
	cout << "No\n";
	exit(0);
}

int main() {
	int N; cin >> N;
	vector<int> num(3);
	for (int i = 0; i < 3; ++i) { cin >> num[i]; }
	vector<string> s(N);
	for (int i = 0; i < N; ++i) { cin >> s[i]; }

	auto wl = [&](int x, int y) {
		++num[x]; --num[y];
	};

	string ABC = "ABC";
	string ans = "";
	for (int i = 0; i < N; ++i) {
		for (int v : num) { if (v < 0) { No(); } }

		int c0 = s[i][0] - 'A', c1 = s[i][1] - 'A';
		if (num[c0] == 0 && num[c1] == 0) { No(); }

		if (num[c0] != num[c1] || i == N - 1) {
			if (num[c0] > num[c1]) {
				wl(c1, c0);
				ans += ABC[c1];
			} else {
				wl(c0, c1);
				ans += ABC[c0];
			}
			continue;
		}
		
		bool add0 = false;
		for (auto c : s[i + 1]) { add0 = add0 || (c == s[i][0]); }
		if (add0) {
			wl(c0, c1);
			ans += ABC[c0];
		} else {
			wl(c1, c0);
			ans += ABC[c1];
		}
	}

	cout << "Yes\n";
	for (auto c : ans) { cout << c << endl; }

	return 0;
}

### Prompt

Generate 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
#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

void No() {
	cout << "No\n";
	exit(0);
}

int main() {
	int N; cin >> N;
	vector<int> num(3);
	for (int i = 0; i < 3; ++i) { cin >> num[i]; }
	vector<string> s(N);
	for (int i = 0; i < N; ++i) { cin >> s[i]; }

	auto wl = [&](int x, int y) {
		++num[x]; --num[y];
	};

	string ABC = "ABC";
	string ans = "";
	for (int i = 0; i < N; ++i) {
		for (int v : num) { if (v < 0) { No(); } }

		int c0 = s[i][0] - 'A', c1 = s[i][1] - 'A';
		if (num[c0] == 0 && num[c1] == 0) { No(); }

		if (num[c0] != num[c1] || i == N - 1) {
			if (num[c0] > num[c1]) {
				wl(c1, c0);
				ans += ABC[c1];
			} else {
				wl(c0, c1);
				ans += ABC[c0];
			}
			continue;
		}
		
		bool add0 = false;
		for (auto c : s[i + 1]) { add0 = add0 || (c == s[i][0]); }
		if (add0) {
			wl(c0, c1);
			ans += ABC[c0];
		} else {
			wl(c1, c0);
			ans += ABC[c1];
		}
	}

	cout << "Yes\n";
	for (auto c : ans) { cout << c << endl; }

	return 0;
}

```