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 <cstdio>
#include <algorithm>
#define maxn 100010
using namespace std;
char in[3];
int n,v[3],p[maxn],q[maxn],ans[maxn];
int main()
{
    int i;
    scanf("%d",&n);
    //ABC
    for(i=0; i<3; i++)
        scanf("%d",&v[i]);
    for(i=1; i<=n; i++)
    {
        scanf("%s",in);
        p[i]=in[0]-'A';
        q[i]=in[1]-'A';
    }
    for(i=1; i<=n; i++)
    {
        if(v[p[i]]>v[q[i]])//小的在前，大的在后
            swap(p[i],q[i]);
        else if(v[p[i]]==v[q[i]])
        {
            if(i+1<=n&&(p[i+1]==q[i]||q[i+1]==q[i]))
                swap(p[i],q[i]);
        }
        if(!v[q[i]])//大的-1都不行了
        {
            return 0*printf("No\n");
        }
        v[p[i]]++;
        v[q[i]]--;
        ans[i]=p[i];//输出加1的那一个字符
    }
    printf("Yes\n");
    for(i=1; i<=n; i++)
    {
        printf("%c\n",ans[i]+'A');
    }
    return 0;
}