#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
	int n, a, b, c;
	scanf("%d %d %d %d", &n, &a, &b, &c);
	map<char, int> cnt;
	cnt['A'] = a;
	cnt['B'] = b;
	cnt['C'] = c;

	string ans;
	vector<string> inp(n);
	for(int i = 0; i < n; i++) {
		char str[4];
		scanf(" %s", str);
		inp[i] = str;
	}
	for(int i = 0; i < n; i++) {
		char t1 = inp[i][0], t2 = inp[i][1];
		if(cnt[t1] == 0 and cnt[t2] == 0) {
			return printf("No\n"), 0;
		}
		if(cnt[t1] < cnt[t2]) {
			swap(t1, t2);
		}
		char t3 = 'A';
		while(t3 == t1 or t3 == t2) t3++;
		if(cnt[t1] == 1 and cnt[t2] == 1 and cnt[t3] == 0) {
			if(i + 1 < n) {
				if(t1 == inp[i+1][0] or t1 == inp[i+1][1]) {
					swap(t1, t2);
				}
			}
		}
		ans += t2;
		cnt[t2]++;
		cnt[t1]--;
	}
	
	printf("Yes\n");
	for(char ch : ans) printf("%c\n", ch);
}

### 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>

using namespace std;
using ll = long long;

int main() {
	int n, a, b, c;
	scanf("%d %d %d %d", &n, &a, &b, &c);
	map<char, int> cnt;
	cnt['A'] = a;
	cnt['B'] = b;
	cnt['C'] = c;

	string ans;
	vector<string> inp(n);
	for(int i = 0; i < n; i++) {
		char str[4];
		scanf(" %s", str);
		inp[i] = str;
	}
	for(int i = 0; i < n; i++) {
		char t1 = inp[i][0], t2 = inp[i][1];
		if(cnt[t1] == 0 and cnt[t2] == 0) {
			return printf("No\n"), 0;
		}
		if(cnt[t1] < cnt[t2]) {
			swap(t1, t2);
		}
		char t3 = 'A';
		while(t3 == t1 or t3 == t2) t3++;
		if(cnt[t1] == 1 and cnt[t2] == 1 and cnt[t3] == 0) {
			if(i + 1 < n) {
				if(t1 == inp[i+1][0] or t1 == inp[i+1][1]) {
					swap(t1, t2);
				}
			}
		}
		ans += t2;
		cnt[t2]++;
		cnt[t1]--;
	}
	
	printf("Yes\n");
	for(char ch : ans) printf("%c\n", ch);
}

```