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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MX = 1e5+2;
string qry[MX];
int cnt[3];
vector<char> ans;
pair<int, int> to_ab(string &s) {
	if(s == "AB")
		return {0, 1};
	else if(s == "AC")
		return {0, 2};
	else
		return {1, 2};
}
int32_t main() {
	ios::sync_with_stdio(0);cin.tie(0);
	int n;
	cin >> n;
	for(int i=0; i<3; i++) {
		cin >> cnt[i];
	}
	for(int i=0; i<n; i++) {
		cin >> qry[i];
	}
	for(int i=0; i<n; i++) {
		int a, b;
		tie(a, b) = to_ab(qry[i]);
		bool add_a = 1;
		if(!cnt[a] && !cnt[b])
			return !(cout << "No");
		else if(cnt[a] == 1 && cnt[b] == 1) {
			auto [na, nb] = to_ab(qry[i+1]);
			add_a = (a == na) || (a == nb);
		} else {
			add_a = cnt[a] < cnt[b];
		}
		if(add_a) {
			ans.push_back(qry[i][0]);
			cnt[a]++;
			cnt[b]--;
			if(cnt[b] < 0)
				exit(123);
		} else {
			ans.push_back(qry[i][1]);
			cnt[b]++;
			cnt[a]--;
			if(cnt[a] < 0)
				exit(456);
		}
	}
	cout << "Yes\n";
	for(char c: ans)
		cout << c << '\n';
}