#include <bits/stdc++.h>
using namespace std;
typedef long long ll ;
const int N=100020;
int v[3],p[N],q[N];
int main(){
    int n;cin >> n;
    string s;
    for(int i=0;i<3;i++)cin >> v[i];
    for(int i=1;i<=n;i++){
        cin >> s;
        p[i]=s[0]-'A';
        q[i]=s[1]-'A';
    }
    for(int i=1;i<=n;i++){
        if(v[p[i]]>v[q[i]])swap(p[i],q[i]);
        else if(v[p[i]]==v[q[i]])
            if(i<n&&(q[i]==q[i+1]||q[i]==p[i+1]))swap(p[i],q[i]);
        if(!v[q[i]])return 0*printf("No\n");
        v[p[i]]++,v[q[i]]--;
    }
    puts("Yes");
    for(int i=1;i<=n;i++) printf("%c\n",'A'+p[i]);
    return 0;
}
### Prompt

Your challenge is to write 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;
typedef long long ll ;
const int N=100020;
int v[3],p[N],q[N];
int main(){
    int n;cin >> n;
    string s;
    for(int i=0;i<3;i++)cin >> v[i];
    for(int i=1;i<=n;i++){
        cin >> s;
        p[i]=s[0]-'A';
        q[i]=s[1]-'A';
    }
    for(int i=1;i<=n;i++){
        if(v[p[i]]>v[q[i]])swap(p[i],q[i]);
        else if(v[p[i]]==v[q[i]])
            if(i<n&&(q[i]==q[i+1]||q[i]==p[i+1]))swap(p[i],q[i]);
        if(!v[q[i]])return 0*printf("No\n");
        v[p[i]]++,v[q[i]]--;
    }
    puts("Yes");
    for(int i=1;i<=n;i++) printf("%c\n",'A'+p[i]);
    return 0;
}
```