B. Weird Rounding

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.

In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, ifk = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.

Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).

It is guaranteed that the answer exists.

Input

The only line of the input contains two integer numbers n and k (0 ≤ n ≤ 2 000 000 000, 1 ≤ k ≤ 9).

It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.

Output

Print w — the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).

Examples

input

30020 3

output

1

input

100 9

output

2

input

10203049 2

output

3

Note

In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.

 

 思路：

　　dfs~~~~；

 

来，上代码：

#include 
         
          #include 
          
           #include 
           
            using namespace std;int n,k,len,ans=0x7fffffff,p;char ch[15];bool if_[15];bool check(){    int bzero=0;    bool zero=true;    long long int pos=0;    for(int i=1;i<=len;i++)    {        if(if_[i]) continue;        if(zero)        {            if(ch[i]=='0') bzero++;            else            {                zero=false;                pos=pos*10+ch[i]-'0';            }        }        else pos=pos*10+ch[i]-'0';    }    if(bzero==1&&pos==0) return true;    if(bzero>1) return false;    if(pos%p==0) return true;    return false;}void dfs(int step,int ci){    if(step>=ans) return ;    if(check())    {        ans=step;        return ;    }    for(int i=ci+1;i<=len;i++)    {        if(!if_[i])        {            if_[i]=true;            dfs(step+1,i);            if_[i]=false;        }    }}int main(){    cin>>ch+1;    cin>>k;    p=1;    for(int i=1;i<=k;i++) p=p*10;    len=strlen(ch+1);    dfs(0,0);    cout<