#include <bits/stdc++.h>
using namespace std;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define RepG(i,u) for(int i=head[u];~i;i=e[i].next)

typedef long long ll;

const int N=1e5+5;

template<typename T> void read(T &x){
   x=0;int f=1;
   char c=getchar();
   for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
   for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
    x*=f;
}

int n,a,b,c;
int ans[N];
char s[N][3];

void dfs(int u,int a,int b,int c){
    if(u>n){
        puts("Yes");
        Rep(i,1,n)printf("%c\n",ans[i]+'A');
        exit(0);
    }
    if(s[u][0]=='A'&&s[u][1]=='B'){
        if(a)ans[u]=1,dfs(u+1,a-1,b+1,c);
        if(b)ans[u]=0,dfs(u+1,a+1,b-1,c);
    }
    if(s[u][0]=='A'&&s[u][1]=='C'){
        if(a)ans[u]=2,dfs(u+1,a-1,b,c+1);
        if(c)ans[u]=0,dfs(u+1,a+1,b,c-1);
    }
    if(s[u][0]=='B'&&s[u][1]=='C'){
        if(b)ans[u]=2,dfs(u+1,a,b-1,c+1);
        if(c)ans[u]=1,dfs(u+1,a,b+1,c-1);
    }
}

int main()
{
    read(n),read(a),read(b),read(c);
    Rep(i,1,n)scanf("%s",s[i]);
    dfs(1,a,b,c);
    puts("No");
    return 0;
}
### 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;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define RepG(i,u) for(int i=head[u];~i;i=e[i].next)

typedef long long ll;

const int N=1e5+5;

template<typename T> void read(T &x){
   x=0;int f=1;
   char c=getchar();
   for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
   for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
    x*=f;
}

int n,a,b,c;
int ans[N];
char s[N][3];

void dfs(int u,int a,int b,int c){
    if(u>n){
        puts("Yes");
        Rep(i,1,n)printf("%c\n",ans[i]+'A');
        exit(0);
    }
    if(s[u][0]=='A'&&s[u][1]=='B'){
        if(a)ans[u]=1,dfs(u+1,a-1,b+1,c);
        if(b)ans[u]=0,dfs(u+1,a+1,b-1,c);
    }
    if(s[u][0]=='A'&&s[u][1]=='C'){
        if(a)ans[u]=2,dfs(u+1,a-1,b,c+1);
        if(c)ans[u]=0,dfs(u+1,a+1,b,c-1);
    }
    if(s[u][0]=='B'&&s[u][1]=='C'){
        if(b)ans[u]=2,dfs(u+1,a,b-1,c+1);
        if(c)ans[u]=1,dfs(u+1,a,b+1,c-1);
    }
}

int main()
{
    read(n),read(a),read(b),read(c);
    Rep(i,1,n)scanf("%s",s[i]);
    dfs(1,a,b,c);
    puts("No");
    return 0;
}
```