#include<bits/stdc++.h>
using namespace std;
int n,a,b,c;
vector<string> v;
#define pb push_back
stack<char> st;
int recur(int idx,int a,int b,int c){
        if(idx==n){return 1; }
        int d[3]; d[0]=a,d[1]=b,d[2]=c;
        d[v[idx][0]-'A']++,d[v[idx][1]-'A']--;
        if(d[0]>=0&&d[1]>=0&&d[2]>=0) { int res=recur(idx+1,d[0],d[1],d[2]); if(res){ st.push(v[idx][0]); return 1; }  }
        d[v[idx][0]-'A']-=2; d[v[idx][1]-'A']+=2;
        if(d[0]>=0&&d[1]>=0&&d[2]>=0) { int res=recur(idx+1,d[0],d[1],d[2]); if(res){ st.push(v[idx][1]); return 1; }  }
        return 0;
}

int main(){
        cin>>n>>a>>b>>c; string s;
        for(int i=1;i<=n;i++) { cin>>s; v.pb(s); }
        int res=recur(0,a,b,c);
        if(!res){ cout<<"No"<<endl; return 0;}
        cout<<"Yes"<<endl;
        while(!st.empty()){
                cout<<st.top()<<endl;
                st.pop();
        }
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int n,a,b,c;
vector<string> v;
#define pb push_back
stack<char> st;
int recur(int idx,int a,int b,int c){
        if(idx==n){return 1; }
        int d[3]; d[0]=a,d[1]=b,d[2]=c;
        d[v[idx][0]-'A']++,d[v[idx][1]-'A']--;
        if(d[0]>=0&&d[1]>=0&&d[2]>=0) { int res=recur(idx+1,d[0],d[1],d[2]); if(res){ st.push(v[idx][0]); return 1; }  }
        d[v[idx][0]-'A']-=2; d[v[idx][1]-'A']+=2;
        if(d[0]>=0&&d[1]>=0&&d[2]>=0) { int res=recur(idx+1,d[0],d[1],d[2]); if(res){ st.push(v[idx][1]); return 1; }  }
        return 0;
}

int main(){
        cin>>n>>a>>b>>c; string s;
        for(int i=1;i<=n;i++) { cin>>s; v.pb(s); }
        int res=recur(0,a,b,c);
        if(!res){ cout<<"No"<<endl; return 0;}
        cout<<"Yes"<<endl;
        while(!st.empty()){
                cout<<st.top()<<endl;
                st.pop();
        }
}

```