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>
#define pb push_back
#define mp make_pair
#define sz(a) a.size()
#define re return
#define all(a) a.begin(),a.end()
#define int long long
using namespace std;
int n, a[3];
string s[100005];
vector<char> ans;
signed main()
{
	cin>>n>>a[0]>>a[1]>>a[2];
	for (int i=0;i<n;i++){
		cin>>s[i];
	}
	for (int i=0;i<n;i++){
		if (!a[s[i][0]-'A'] && !a[s[i][1]-'A']){
			cout<<"No";
			re 0;
		}
		else if (a[s[i][0]-'A']>a[s[i][1]-'A']){
			ans.pb(s[i][1]);
			a[s[i][0]-'A']--;
			a[s[i][1]-'A']++;
		}
		else if (a[s[i][0]-'A']<a[s[i][1]-'A']){
			ans.pb(s[i][0]);
			a[s[i][0]-'A']++;
			a[s[i][1]-'A']--;
		}
		else if (a[s[i][0]-'A']==a[s[i][1]-'A']){
			if (i==n-1 || s[i+1][0]==s[i][0] || s[i+1][1]==s[i][0]){
				ans.pb(s[i][0]);
				a[s[i][0]-'A']++;
				a[s[i][1]-'A']--;
			}
			else{
				ans.pb(s[i][1]);
				a[s[i][0]-'A']--;
				a[s[i][1]-'A']++;
			}
		}
		//cout<<a[0]<<' '<<a[1]<<' '<<a[2]<<endl;
	}
	cout<<"Yes"<<endl;
	for (int i=0;i<n;i++)
		cout<<ans[i]<<endl;
	re 0;
}