#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <string>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> PLL;

const ll N = 100010;
ll n, a[3];
string s[N];
string res;

ll get(string s) {
	if (s == "AB") return 0;
	if (s == "BC") return 1;
	if (s == "AC") return 2;
}

int main() {
	cin >> n;
	ll sum = 0;
	for (ll i = 0; i < 3; i++) {
		cin >> a[i];
		sum += a[i];
	}

	for (ll i = 1; i <= n; i++) {
		cin >> s[i];
	}

	for (ll i = 1; i <= n; i++) {
		ll x = get(s[i]), y = (x + 1) % 3;
		ll p, q;

		if (a[x] >= 2) {
			p = x;
		}
		else if (a[y] >= 2) {
			p = y;
		}
		else if (a[x] == 1 && a[y] == 1) {
			if (i == n || get(s[i + 1]) == y) {
				p = x;
			}
			else {
				p = y;
			}
		}
		else if (a[x] == 1) {
			p = x;
		}
		else if (a[y] == 1) {
			p = y;
		}
		else {
			cout << "No" << endl;
			return 0;
		}

		if (p == x) {
			q = y;
		}
		else {
			q = x;
		}

		a[p]--, a[q]++;
		res += "ABC"[q];
	}

	cout << "Yes" << endl;
	for (int i = 0; i < res.size(); i++) {
		cout << res[i] << endl;
	}

	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <string>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> PLL;

const ll N = 100010;
ll n, a[3];
string s[N];
string res;

ll get(string s) {
	if (s == "AB") return 0;
	if (s == "BC") return 1;
	if (s == "AC") return 2;
}

int main() {
	cin >> n;
	ll sum = 0;
	for (ll i = 0; i < 3; i++) {
		cin >> a[i];
		sum += a[i];
	}

	for (ll i = 1; i <= n; i++) {
		cin >> s[i];
	}

	for (ll i = 1; i <= n; i++) {
		ll x = get(s[i]), y = (x + 1) % 3;
		ll p, q;

		if (a[x] >= 2) {
			p = x;
		}
		else if (a[y] >= 2) {
			p = y;
		}
		else if (a[x] == 1 && a[y] == 1) {
			if (i == n || get(s[i + 1]) == y) {
				p = x;
			}
			else {
				p = y;
			}
		}
		else if (a[x] == 1) {
			p = x;
		}
		else if (a[y] == 1) {
			p = y;
		}
		else {
			cout << "No" << endl;
			return 0;
		}

		if (p == x) {
			q = y;
		}
		else {
			q = x;
		}

		a[p]--, a[q]++;
		res += "ABC"[q];
	}

	cout << "Yes" << endl;
	for (int i = 0; i < res.size(); i++) {
		cout << res[i] << endl;
	}

	return 0;
}
```