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:

// Problem : F - Three Variables Game
// Contest : AtCoder - AtCoder Beginner Contest 166
// URL : https://atcoder.jp/contests/abc166/tasks/abc166_f
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>

using namespace std;

string ss[100005];

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int a[3];
	int N;
	cin >> N >> a[0] >> a[1] >> a[2];
	vector<char> ans;
	for(int i = 1; i<=N; i++){
		cin >> ss[i];
	}
	ss[N+1] = "AB";
	for(int i = 1; i<=N; i++){
		string s = ss[i];
		if(a[s[0]-'A'] == 0 && a[s[1]-'A'] == 0){
			cout << "No\n";
			return 0;
		}
		else if(a[s[0]-'A'] == a[s[1]-'A']){
			if(ss[i+1][0] == s[1] || ss[i+1][1] == s[1]){
				a[s[1]-'A']++, a[s[0]-'A']--;
				ans.push_back(s[1]);
			}
			else{
				a[s[1]-'A']--, a[s[0]-'A']++;
				ans.push_back(s[0]);
			}
		}
		else if(a[s[0]-'A'] > a[s[1]-'A']){
			ans.push_back(s[1]);
			a[s[0]-'A']--, a[s[1]-'A']++;
		}
		else{
			ans.push_back(s[0]);
			a[s[0]-'A']++, a[s[1]-'A']--;
		}
	}
	cout << "Yes\n";
	for(auto c : ans){
		cout << c << "\n";
	}
}