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;
typedef long long ll;
const int N=1e5+5;
string s[N];
char ans[N];
int n,a,b,c;
void dfs(int id,int a,int b,int c){
    if(a<0 || b<0 || c<0) return;
    if(id==n){
        puts("Yes");
        for(int i=0;i<n;i++) cout<<ans[i]<<endl;
        exit(0);
    }
    else{
        if(s[id]=="AB"){
            ans[id]='A';
            dfs(id+1,a+1,b-1,c);
            ans[id]='B';
            dfs(id+1,a-1,b+1,c);
        }
        else if(s[id]=="AC"){
            ans[id]='A';
            dfs(id+1,a+1,b,c-1);
            ans[id]='C';
            dfs(id+1,a-1,b,c+1);
        }
        else if(s[id]=="BC"){
            ans[id]='B';
            dfs(id+1,a,b+1,c-1);
            ans[id]='C';
            dfs(id+1,a,b-1,c+1);
        }
    }
}
int main()
{
    cin>>n>>a>>b>>c;
    for(int i=0;i<n;i++) cin>>s[i];
    dfs(0,a,b,c);
    puts("No");
    return 0;
}