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.


# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def ri():
	return int(f.readline().strip())
def ria():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def rs():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]





from collections import deque
import math
# NUMBER = 10**9 + 7
NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
def choose(n,k):
	if n < k:
		assert false
	return  mult(factorial(n), modinv(mult(factorial(k),factorial(n-k))))
from collections import deque, defaultdict 
import heapq


from types import GeneratorType
def bootstrap(f, stack=[]):
	def wrappedfunc(*args, **kwargs):
		if stack:
			return f(*args, **kwargs)
		else:
			to = f(*args, **kwargs)
			while True:
				if type(to) is GeneratorType:
					stack.append(to)
					to = next(to)
				else:
					stack.pop()
					if not stack:
						break
					to = stack[-1].send(to)
			return to

	return wrappedfunc


def dfs(g, timeIn, timeOut,depths,parents): 
	# assign In-time to node u 
	cnt = 0
	# node, neig_i, parent, depth
	stack = [[1,0,0,0]]
	while stack:
		v,neig_i,parent,depth = stack[-1]
		parents[v] = parent
		depths[v] = depth
		# print (v)
		if neig_i == 0:
			timeIn[v] = cnt    
			cnt+=1
		while neig_i < len(g[v]):
			u = g[v][neig_i]
			if u == parent:
				neig_i+=1
				continue
			stack[-1][1] = neig_i + 1
			stack.append([u,0,v,depth+1])
			break
		if neig_i == len(g[v]):
			stack.pop()
			timeOut[v] = cnt
			cnt += 1
  
def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str: 
	return timeIn[u] <= timeIn[v] and timeOut[v] <= timeOut[u] 

cnt = 0
@bootstrap
def dfs(v,adj,timeIn, timeOut,depths,parents,parent=0,depth=0):
	global cnt
	parents[v] = parent
	depths[v] = depth
	timeIn[v] = cnt
	cnt+=1
	for u in adj[v]:
		if u == parent:
			continue
		yield dfs(u,adj,timeIn,timeOut,depths,parents,v,depth+1)
	timeOut[v] = cnt
	cnt+=1
	yield 


def solution(s,n,k):
	a = []
	sol = 0
	diff = 0
	seen_one = False
	for i in range(n):
		x = s[i]
		if x == '0':
			diff+=1
			continue
		if not seen_one:
			seen_one=True
			sol+=diff//(k+1)
		else:
			sol+=max(0,(diff-k)//(k+1))
		diff = 0
	# print(sol,diff)
	return sol + (diff//(k+1) if seen_one else  ((diff-1)//(k+1) + 1))



	
def main():
	T = 1
	T = ri()
	for i in range(T):
		# n = ri()
		n,k = ria()
		s = rs()
		# a=ria()

		x = solution(s,n,k)
		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()

