Most people know that the binary operations. Do you know the binary mod and divide?

Now give the Binary number N and a integer number M ,Can you tell me the answer of N%(2^M) and N/(2^M)?

输入

Input contains multiple test cases.

The first line of each test case contains an binary number N no more than 128 bits and an integer M (1 <= M <= 64).

when N=0&&M=0 ,test is over.

输出

output the answer the N%(2^M) and N/(2^M).

样例输入

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 using namespace std;

 char a[];

 void add(int l,int r,int b[])

 {

     for(int i=l;i<r;i++)

     {

         for(int j=;j<;j++)

             b[j]*=;

         b[]+=a[i]-'';

         for(int j=;j<;j++)

         {

             if(b[j]>=)

             {

                 b[j+]+=b[j]/;

                 b[j]%=;

             }

         }

     }

 }

 int main()

 {

     int m,i,l;

     int b[],c[];

     while(scanf("%s%d",a,&m))

     {

         if(!m&&a[]=='') break;

         memset(b,,sizeof(b));

         memset(c,,sizeof(c));

         l=strlen(a);

         add(,l-m,b);

         add(max(,l-m),l,c);
35         printf("mod=");
         for(i=;i>=;i--)

             if(c[i]) break;

         if(i==-) printf("");

         for(;i>=;i--)

             printf("%d",c[i]);
41　　　　　 printf(", divide=");
         for(i=;i>=;i--)

             if(b[i]) break;

         if(i==-) printf("");

         for(;i>=;i--)

             printf("%d",b[i]);

         putchar();

     }

 }