#include <bits/stdc++.h>
#define rep(i,n) for(ll i = 0; i < (ll)(n); ++i)
#define MOD 1000000007

using namespace std;
using ll = long long;

int main() {
	int n, h[3];
	string ans;
	cin >> n;
	rep(i,3) cin >> h[i];
	vector<pair<int, int>> m;
	rep(i, n) {
		string str;
		cin >> str;
		m.emplace_back(str[0] - 'A', str[1] - 'A');
	}
	auto play = [&] (int x, int y) {
		h[x]++;
		h[y]--;
		ans += 'A' + x;
		return;
	};
	m.emplace_back(0, 1);
	rep(i, n) {
		int s = m[i].first;
		int t = m[i].second;
		if (h[s] + h[t] == 0) {
			cout << "No" << endl;
			return 0;
		}
		if (h[s] == h[t]) {
			int ns = m[i + 1].first;
			int nt = m[i + 1].second;
			if (s == ns || s == nt) play(s, t);
			else play(t, s);
		}
		else if (h[s] > h[t]) play(t, s);
		else play(s, t);
	}
	cout << "Yes" << endl;
	for (char 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 rep(i,n) for(ll i = 0; i < (ll)(n); ++i)
#define MOD 1000000007

using namespace std;
using ll = long long;

int main() {
	int n, h[3];
	string ans;
	cin >> n;
	rep(i,3) cin >> h[i];
	vector<pair<int, int>> m;
	rep(i, n) {
		string str;
		cin >> str;
		m.emplace_back(str[0] - 'A', str[1] - 'A');
	}
	auto play = [&] (int x, int y) {
		h[x]++;
		h[y]--;
		ans += 'A' + x;
		return;
	};
	m.emplace_back(0, 1);
	rep(i, n) {
		int s = m[i].first;
		int t = m[i].second;
		if (h[s] + h[t] == 0) {
			cout << "No" << endl;
			return 0;
		}
		if (h[s] == h[t]) {
			int ns = m[i + 1].first;
			int nt = m[i + 1].second;
			if (s == ns || s == nt) play(s, t);
			else play(t, s);
		}
		else if (h[s] > h[t]) play(t, s);
		else play(s, t);
	}
	cout << "Yes" << endl;
	for (char c : ans) cout << c << endl;
	return 0;
}

```