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;
using ll = long long;
int n,a,b,c;
string ans;
vector<string> s;
map<pair<pair<int,int>,pair<int,int>>,int> mp;
using P = pair<int,int>;
void dfs(int now,int A,int B,int C)
{
	if(mp[pair<P,P>(P(now,A),P(B,C))]==0)
	{
		mp[pair<P,P>(P(now,A),P(B,C))]=1;
	}
	else
	{
		return;
	}
	
	if(A==-1||B==-1||C==-1)
	{
		return;
	}
	if(now==n)
	{
		cout<<"Yes"<<endl;
		for(auto c:ans)
		{
			cout<<c<<endl;
		}
		exit(0);
	}

	if(s[now]=="AB")
	{
		ans[now]='B';
		dfs(now+1,A-1,B+1,C);
		ans[now]='A';
		dfs(now+1,A+1,B-1,C);
	}
	if(s[now]=="BC")
	{
		ans[now]='B';
		dfs(now+1,A,B+1,C-1);
		ans[now]='C';
		dfs(now+1,A,B-1,C+1);
	}
	if(s[now]=="AC")
	{
		ans[now]='A';
		dfs(now+1,A+1,B,C-1);
		ans[now]='C';
		dfs(now+1,A-1,B,C+1);
	}
}

int main()
{
	cin >> n >> a >>b >> c;
	for(int i = 0;i<n;i++)
	{
		string t;
		cin >> t;
		s.push_back(t);
	}
	ans = string(n,0);
	dfs(0,a,b,c);
	cout<<"No"<<endl;
}