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;
#include<cstdio>
#define el "\n"
#define sp " "
typedef long long ll;

vector<string> ans(100005);
vector<string> s(100005);
ll n,y=0;
void dfs(ll i,ll a,ll b, ll c)
{
	if(y)
	return ;
	if(i==n)
	{
		cout<<"Yes"<<el;
		for(ll j=0;j<n;j++)
		cout<<ans[j]<<el;
		y=1;
	}
	if(s[i]=="AB")
	{
		if(b) ans[i]="A",dfs(i+1,a+1,b-1,c);
		if(a) ans[i]="B",dfs(i+1,a-1,b+1,c);
		
	}
		if(s[i]=="AC")
	{
		if(c) ans[i]="A",dfs(i+1,a+1,b,c-1);
		if(a) ans[i]="C",dfs(i+1,a-1,b,c+1);
	}
		if(s[i]=="BC")
	{
			if(c) ans[i]="B",dfs(i+1,a,b+1,c-1);
			if(b) ans[i]="C",dfs(i+1,a,b-1,c+1);
	}
	
}
int main()
{
	ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll a,b,c;
cin>>n>>a>>b>>c;
for(ll i=1;i<=n;i++)
cin>>s[i-1];
dfs(0,a,b,c);
if(!y)
cout<<"No";

}