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 java.io.*;
import java.util.*;

public class Main {
	
	public static FastReader scan = new FastReader();

	public static void main(String[] args) {
		
		int t = scan.nextInt();
		
		while(t--> 0) {
			int n = scan.nextInt();
			int k = scan.nextInt();
			String s = scan.next();
			int ans = 0;
			
			for(int i=0; i<n;) {
				int j = i+1;
				
				for(; j<n && s.charAt(j) != '1'; j++);
				
				int left = s.charAt(i) == '1' ? k : 0;
				int right = j < n && s.charAt(j) == '1' ? k : 0;
				int len = j - i;
				
				if(left == k) len--;
				
				len -= (left + right);
				
				if(len > 0) ans += (len + k)/(k+1);
				
				i = j;
				
			}
			
			System.out.println(ans);
			
		}
		
  }

	public static int gcd(int a, int b) {
		if(b == 0) return a;
		return gcd(b, a%b);
	}

	public static long abs(long a, long b) {return Math.abs(a - b);}
	public static long min(long a, long b) {return Math.min(a, b);}
	public static long max(long a, long b) {return Math.max(a, b);}
	
	public static int abs(int a, int b) {return Math.abs(a - b);}
	public static int min(int a, int b) {return Math.min(a, b);}
	public static int max(int a, int b) {return Math.max(a, b);}
	
	// sieveOfEratosthenes 
	public static boolean[] sieve(int n) {
		boolean[] prime = new boolean[n+1];
		
		Arrays.fill(prime, true);
		prime[0] = false;
		prime[1] = false;
		
		for(int i=2; i*i<=n; i++) {
			if(prime[i] == true) {				
				for(int j=i*2; j<=n; j+=i) {
					prime[j] = false;
				}
			}
		}
		
		return prime;
	}
	
	//smallestFactor function
	public static int smallestFactor(int n) {
		int i;
		for(i=2; i*i<=n; i++) {
			if(n%i == 0) 
				break;
		}
		return i;
	}
	
	//isPrime 
	public static boolean isPrime(long n) {
		boolean isPrime = true;
		for(int i=2; i*i<=n; i++) if(n % i == 0) isPrime = false;
		return isPrime;
	}

	//FastReader Class
	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastReader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 

		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 
		
		char[] nextChar() 
		{
			return next().toCharArray();
		}

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	} 
	
	private static long gcd(long m, long n) {
		if(isINFL(-m)) return n;
		assertion((m>=0L) && (n>=0L));
		if(m < n) return gcd(n, m);
		if(n == 0) return m;
		return gcd(n, m % n);
	}
	

	private static void assertion(boolean b) {
		if(!b) throw new AssertionError();
	}

	private static long INFL = (long)1e17;
	private static boolean isINFL(long in) {
		if((in*10L)>INFL) return true;
		else return false;
	}
	
}