C. Social Distance

C. Social Distance

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp and his friends want to visit a new restaurant. The restaurant has nn tables arranged along a straight line. People are already sitting at some tables. The tables are numbered from 11 to nn in the order from left to right. The state of the restaurant is described by a string of length nn which contains characters "1" (the table is occupied) and "0" (the table is empty).

Restaurant rules prohibit people to sit at a distance of kk or less from each other. That is, if a person sits at the table number ii, then all tables with numbers from i−ki−k to i+ki+k (except for the ii-th) should be free. In other words, the absolute difference of the numbers of any two occupied tables must be strictly greater than kk.

For example, if n=8n=8 and k=2k=2, then:

• strings "10010001", "10000010", "00000000", "00100000" satisfy the rules of the restaurant;
• strings "10100100", "10011001", "11111111" do not satisfy to the rules of the restaurant, since each of them has a pair of "1" with a distance less than or equal to k=2k=2.

In particular, if the state of the restaurant is described by a string without "1" or a string with one "1", then the requirement of the restaurant is satisfied.

You are given a binary string ss that describes the current state of the restaurant. It is guaranteed that the rules of the restaurant are satisfied for the string ss.

Find the maximum number of free tables that you can occupy so as not to violate the rules of the restaurant. Formally, what is the maximum number of "0" that can be replaced by "1" such that the requirement will still be satisfied?

For example, if n=6n=6, k=1k=1, s=s= "100010", then the answer to the problem will be 11, since only the table at position 33 can be occupied such that the rules are still satisfied.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases in the test. Then tt test cases follow.

Each test case starts with a line containing two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of tables in the restaurant and the minimum allowed distance between two people.

The second line of each test case contains a binary string ss of length nn consisting of "0" and "1" — a description of the free and occupied tables in the restaurant. The given string satisfy to the rules of the restaurant — the difference between indices of any two "1" is more than kk.

The sum of nn for all test cases in one test does not exceed 2⋅1052⋅105.

Output

For each test case output one integer — the number of tables that you can occupy so as not to violate the rules of the restaurant. If additional tables cannot be taken, then, obviously, you need to output 00.

Example

input

Copy

6
6 1
100010
6 2
000000
5 1
10101
3 1
001
2 2
00
1 1
0

output

Copy

1
2
0
1
1
1

Note

The first test case is explained in the statement.

In the second test case, the answer is 22, since you can choose the first and the sixth table.

In the third test case, you cannot take any free table without violating the rules of the restaurant.

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贪心，从前往后遍历即可，能加就加，加的条件是左右k距离没有1

判断区间内是否有1的方法是前缀和，但本题是动态修改的，也就是当前位置前面是否有1是需要根据之前修改情况也判断的，当然，右边是否有1是完全可以利用前缀和来判断的，至于前面的，我们设置一个pre变量即可，每当遇见1就修改，值得注意的是，我们新添进去的1也不能忘记修改pre

#include<iostream>
#include<cstdio>
#include<cstring>
# include<iomanip>

#include<algorithm>
#define  mo 998244353;
using namespace std;

typedef long long int ll;


int sum[200000+10];
int main()
{


     int t;

     cin>>t;

     while(t--)
     {
         string s;

         int n,k;

         cin>>n>>k;

         cin>>s;
         s=" "+s;

         for(int i=1;i<=n;i++)
         {
             sum[i]=sum[i-1]+s[i]=='1';
         }
         int pre=-1e9;

         int ans=0;
         for(int i=1;i<=n;i++)
         {
             if(s[i]=='1')
             {
                 pre=i;
             }
             else if(s[i]=='0')
             {

                 if(pre+k<i&&sum[min(n,i+k)]-sum[i]==0)
                 {

                     ans++;

                     pre=i;
                 }
             }
         }

         cout<<ans<<endl;




     }
    return 0;
}