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;
const int N = 1e5 + 5;
string s[N];
int ok = 1;
vector<char> ans;
inline void op(int &a,int &b,char ans1,char ans2,bool jug) {
	if(a == 0 && b == 0) {ok = 0;return;}
	if(a == 1 && b == 1) {
		if(jug == 1) --a,++b,ans.push_back(ans2);
		else ++a,--b,ans.push_back(ans1);
	}
	else if(a < b) --b,++a,ans.push_back(ans1);
	else --a,++b,ans.push_back(ans2);
}
int main() {
	ios::sync_with_stdio(0);
	int n,a,b,c;
	cin >> n >> a >> b >> c;
	for(int i = 1;i <= n;++i)
		cin >> s[i];
	s[n+1] = "DD";
	for(int i = 1;i <= n;++i) {
		if(s[i][0] == 'A' && s[i][1] == 'B') op(a,b,'A','B',c == 0 && s[i+1] == "BC");
		if(s[i][0] == 'A' && s[i][1] == 'C') op(a,c,'A','C',b == 0 && s[i+1] == "BC");
		if(s[i][0] == 'B' && s[i][1] == 'C') op(b,c,'B','C',a == 0 && s[i+1] == "AC");
		if(ok == 0) break;
	}
	if(!ok) cout << "No" << endl;
	else {
		cout <<"Yes" << endl;
		for(int i = 0;i < n;++i)
			cout << ans[i] << endl;
	}
	return 0;
}