#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	int n; cin >> n;
	vector<int> A(3);
	vector<string> s(n);
	vector<char> ans;
	for (auto& i : A) cin >> i;
	for (auto& i : s) cin >> i;
	for (int i = 0; i < n; ++i) {
		int a = s[i][0] - 'A', b = s[i][1] - 'A';
		if (!A[a] && !A[b]) {
			cout << "No";
			return 0;
		}
		else if (A[a] > A[b]) {
			ans.push_back('A' + b);
			A[a]--, A[b]++;
		}
		else if (A[a] < A[b]) {
			ans.push_back('A' + a);
			A[a]++, A[b]--;
		}
		else {
			if (i == n - 1 || s[i] == s[i + 1]) {
				ans.push_back('A' + a);
				A[b]--; A[a]++;
			}
			else {
				int c = s[i + 1][0] - 'A', d = s[i + 1][1] - 'A';
				if (a == c || a == d) {
					ans.push_back('A' + a);
					A[b]--; A[a]++;
				}
				else {
					ans.push_back('A' + b);
					A[a]--; A[b]++;
				}
			}
		}
	}
	cout << "Yes\n";
	for (auto i : ans) cout << i << '\n';
}
### Prompt

Create a solution in Cpp for 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 ll long long
using namespace std;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	int n; cin >> n;
	vector<int> A(3);
	vector<string> s(n);
	vector<char> ans;
	for (auto& i : A) cin >> i;
	for (auto& i : s) cin >> i;
	for (int i = 0; i < n; ++i) {
		int a = s[i][0] - 'A', b = s[i][1] - 'A';
		if (!A[a] && !A[b]) {
			cout << "No";
			return 0;
		}
		else if (A[a] > A[b]) {
			ans.push_back('A' + b);
			A[a]--, A[b]++;
		}
		else if (A[a] < A[b]) {
			ans.push_back('A' + a);
			A[a]++, A[b]--;
		}
		else {
			if (i == n - 1 || s[i] == s[i + 1]) {
				ans.push_back('A' + a);
				A[b]--; A[a]++;
			}
			else {
				int c = s[i + 1][0] - 'A', d = s[i + 1][1] - 'A';
				if (a == c || a == d) {
					ans.push_back('A' + a);
					A[b]--; A[a]++;
				}
				else {
					ans.push_back('A' + b);
					A[a]--; A[b]++;
				}
			}
		}
	}
	cout << "Yes\n";
	for (auto i : ans) cout << i << '\n';
}
```