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

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

For example, if n=8 and k=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=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 s that describes the current state of the restaurant. It is guaranteed that the rules of the restaurant are satisfied for the string s.

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=6, k=1, s= "100010", then the answer to the problem will be 1, since only the table at position 3 can be occupied such that the rules are still satisfied.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.

Each test case starts with a line containing two integers n and k (1 ≤ k ≤ n ≤ 2⋅ 10^5) — 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 s of length n 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 k.

The sum of n for all test cases in one test does not exceed 2⋅ 10^5.

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 0.

Example

Input


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


Output


1
2
0
1
1
1

Note

The first test case is explained in the statement.

In the second test case, the answer is 2, 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.
import com.sun.org.apache.bcel.internal.generic.ARRAYLENGTH;

import javax.print.DocFlavor;
import javax.print.attribute.standard.RequestingUserName;
import java.io.*;
import java.lang.reflect.Array;
import java.net.Inet4Address;
import java.nio.charset.IllegalCharsetNameException;
import java.time.temporal.Temporal;
import java.time.temporal.TemporalAccessor;
import java.util.*;
import java.util.concurrent.locks.ReentrantLock;

public class Solution {
    public static void main(String[] args) throws IOException{
        Reader.init(System.in);
        long h = 973303; long c  = 519683 ; long t = 866255;
        double lol = (double)(h-t) /(double) (2*t - h - c);
//        System.out.println(lol);
        t =Reader.nextInt();
//        t =1;
//        System.out.println(cnt);
        PrintWriter out = new PrintWriter(System.out);
        int i = 1;
        while (t-->0){
//            out.print("Case #" + i + ": ");
            solve(out);
            i++;
        }
        out.close();
    }
    static class Pair{
        int l , r;

        public Pair(int l, int r) {
            this.l = l;
            this.r = r;
        }
    }
    static int[][] arr;
    static int n;
    static int m;
    static int[][] dep1;
    static int[][] dep2;
    static boolean[][] v1;
    static boolean[][] v2;
    static TreeSet<Integer> set = new TreeSet<>();
    static void solve(PrintWriter out)throws IOException {
        int n= Reader.nextInt();
        int k = Reader.nextInt();
        char[] arr = Reader.next().toCharArray();
        TreeSet<Integer> set = new TreeSet<>();
        for (int i = 0 ; i < n ; i++){
            if (arr[i]=='1'){
                set.add(i);
            }
        }
        int c = 0;
        for (int i = 0 ; i < n ; i++){
            if (arr[i]=='1'){
                continue;
            }
            Integer l = set.lower(i);
            Integer r = set.higher(i);
//            System.out.println(set.toString());
//            System.out.println(l + " " + r + " "+ i);
            if (l!=null){
//                System.out.println(i-l);
                if (i-l<=k){
                    continue;
                }
            }
            if (r!=null){
                if (r-i<=k){
                    continue;
                }
            }
            arr[i]='1';
            set.add(i);
            c++;
        }
        out.println(c);
    }

    static void sieveOfEratosthenes(int n)
    {
        // Create a boolean array "prime[0..n]" and initialize
        // all entries it as true. A value in prime[i] will
        // finally be false if i is Not a prime, else true.
        boolean prime[] = new boolean[n+1];
        for(int i=0;i<n;i++)
            prime[i] = true;

        for(int p = 2; p*p <=n; p++)
        {
            // If prime[p] is not changed, then it is a prime
            if(prime[p] == true)
            {
                // Update all multiples of p
                for(int i = p*p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        // Print all prime numbers
        for(int i = 2; i <= n; i++)
        {
            if(prime[i] == true){
                set.add(i);
            }
        }
    }

    static void dfs1(int i , int j, int d){
        if (check(i,j)){
            if (v1[i][j]){
                return;
            }
            v1[i][j] = true;
//            System.out.println(i+" " + j + " " + d);
            dep1[d][arr[i][j]]++;
            dfs1(i+1,j,d+1);
            dfs1(i,j+1 , d+1);
        }
        else{
            return;
        }
    }
    static void dfs2(int i , int j, int d){
        if (check(i,j)){
            if (v2[i][j]){
                return;
            }
            v2[i][j] = true;
            dep1[d][arr[i][j]]++;
            dfs2(i-1,j,d+1);
            dfs2(i,j-1 , d+1);
        }
        else{
            return;
        }
    }
    static boolean check(int i , int j ){
        if (i<0 || j <0 || i>= n || j>=m){
            return false;
        }
        return true;
    }


    public static int stringCompare(String str1, String str2)
    {

        int l1 = str1.length();
        int l2 = str2.length();
        int lmin = Math.min(l1, l2);

        for (int i = 0; i < lmin; i++) {
            int str1_ch = (int)str1.charAt(i);
            int str2_ch = (int)str2.charAt(i);

            if (str1_ch != str2_ch) {
                return str1_ch - str2_ch;
            }
        }

        // Edge case for strings like
        // String 1="Geeks" and String 2="Geeksforgeeks"
        if (l1 != l2) {
            return l1 - l2;
        }

        // If none of the above conditions is true,
        // it implies both the strings are equal
        else {
            return 0;
        }
    }








    static int next(long[] arr, long target)
    {
        int start = 0, end = arr.length - 1;

        int ans = -1;
        while (start <= end) {
            int mid = (start + end) / 2;

            // Move to right side if target is
            // greater.
            if (arr[mid] <= target) {
                start = mid + 1;
            }

            // Move left side.
            else {
                ans = mid;
                end = mid - 1;
            }
        }
        return ans;
    }

//    static int find(int x) {
//        return parent[x] == x ? x : (parent[x] = find(parent[x]));
//    }
//    static boolean merge(int x, int y) {
//        x = find(x);
//        y = find(y);
//        if (x==y){
//            return false;
//        }
//        parent[x] = y;
//        return true;
//    }







    public static void sortbyColumn(int arr[][], int col)
    {
        // Using built-in sort function Arrays.sort
        Arrays.sort(arr, new Comparator<int[]>() {

            @Override
            // Compare values according to columns
            public int compare(final int[] entry1,
                               final int[] entry2) {

                // To sort in descending order revert
                // the '>' Operator
                if (entry1[col] > entry2[col])
                    return 1;
                else
                    return -1;
            }
        });  // End of function call sort().
    }



}


class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}
class Pair{
    int a;
    int b;

    public Pair(int a, int b) {
        this.a = a;
        this.b = b;
    }
}
class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]

    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    /* A utility function to print array of size n */
    static void printArray(int arr[])
    {
        int n = arr.length;
        for (int i=0; i<n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    // Driver method

}

class Node implements Comparable<Node>{
    int a;
    int b;
    Node (int a , int b){
        this.a = a;
        this.b = b;
    }



    public int compareTo(Node o) {
        if ((this.a%2) == (o.a%2)){
            return (this.b - o.b);
        }
        else{
            return this.a - o.a;
        }
    }
}
class Graph
{
    int V;// No. of vertices

    //An Array of List which contains
    //references to the Adjacency List of
    //each vertex
    List <Integer> adj[];
    public Graph(int V)// Constructor
    {
        this.V = V;
        adj = new ArrayList[V];
        for(int i = 0; i < V; i++)
            adj[i]=new ArrayList<Integer>();
    }

    // function to add an edge to graph
    public void addEdge(int u,int v)
    {
        adj[u].add(v);
    }
    // prints a Topological Sort of the complete graph
    public void topologicalSort(PrintWriter out,TreeSet set)
    {
        // Create a array to store indegrees of all
        // vertices. Initialize all indegrees as 0.
        int indegree[] = new int[V];

        // Traverse adjacency lists to fill indegrees of
        // vertices. This step takes O(V+E) time
        for(int i = 0; i < V; i++)
        {
            ArrayList<Integer> temp = (ArrayList<Integer>) adj[i];
            for(int node : temp)
            {
                indegree[node]++;
            }
        }

        // Create a queue and enqueue all vertices with
        // indegree 0
        Queue<Integer> q = new LinkedList<Integer>();
        for(int i = 0;i < V; i++)
        {
            if(indegree[i]==0)
                q.add(i);
        }

        // Initialize count of visited vertices
        int cnt = 0;

        // Create a vector to store result (A topological
        // ordering of the vertices)
        Vector <Integer> topOrder=new Vector<Integer>();
        while(!q.isEmpty())
        {
            // Extract front of queue (or perform dequeue)
            // and add it to topological order
            int u=q.poll();
            topOrder.add(u);

            // Iterate through all its neighbouring nodes
            // of dequeued node u and decrease their in-degree
            // by 1
            for(int node : adj[u])
            {
                // If in-degree becomes zero, add it to queue
                if(--indegree[node] == 0)
                    q.add(node);
            }
            cnt++;
        }

        // Check if there was a cycle
        if(cnt != V)
        {
            out.println(-1);
            return ;
        }

        // Print topological order
        for(int i : topOrder)
        {
            if (set.contains((char)('A'+i))) {
                out.print((char) ('A' + i));
            }
        }
        out.println();
    }
}
// Driver program to test above functions