#include<iostream>
#include<vector>
#include<string>
#define rip(i,n,s) for(int i=(s);i<(int)(n);i++)
using namespace std;
int g[3],ans[100000];
string d[100001];
void j(int i,int x,int y){ans[i]=x;g[x]++;g[y]--;}
int main(){
    int n;cin>>n>>g[0]>>g[1]>>g[2];
    rip(i,n,0)cin>>d[i];
    d[n]="AB";
    rip(i,n,0){
        int x=d[i][0]-'A';
        int y=d[i][1]-'A';
        if(g[x]==g[y]){
            if(d[i+1][0]==d[i][0]||d[i+1][1]==d[i][0])j(i,x,y);
            else j(i,y,x);
            if(g[x]<0||g[y]<0){printf("No\n");return 0;}
        }
        else if(g[x]>g[y])j(i,y,x);
        else j(i,x,y);
    }
    printf("Yes\n");
    rip(i,n,0) printf("%c\n",ans[i]+'A');
}
### Prompt

Please create a solution in Cpp 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<iostream>
#include<vector>
#include<string>
#define rip(i,n,s) for(int i=(s);i<(int)(n);i++)
using namespace std;
int g[3],ans[100000];
string d[100001];
void j(int i,int x,int y){ans[i]=x;g[x]++;g[y]--;}
int main(){
    int n;cin>>n>>g[0]>>g[1]>>g[2];
    rip(i,n,0)cin>>d[i];
    d[n]="AB";
    rip(i,n,0){
        int x=d[i][0]-'A';
        int y=d[i][1]-'A';
        if(g[x]==g[y]){
            if(d[i+1][0]==d[i][0]||d[i+1][1]==d[i][0])j(i,x,y);
            else j(i,y,x);
            if(g[x]<0||g[y]<0){printf("No\n");return 0;}
        }
        else if(g[x]>g[y])j(i,y,x);
        else j(i,x,y);
    }
    printf("Yes\n");
    rip(i,n,0) printf("%c\n",ans[i]+'A');
}
```