You are given an array a[0 … n - 1] = [a_0, a_1, …, a_{n - 1}] of zeroes and ones only. Note that in this problem, unlike the others, the array indexes are numbered from zero, not from one.

In one step, the array a is replaced by another array of length n according to the following rules: 

  1. First, a new array a^{→ d} is defined as a cyclic shift of the array a to the right by d cells. The elements of this array can be defined as a^{→ d}_i = a_{(i + n - d) mod n}, where (i + n - d) mod n is the remainder of integer division of i + n - d by n. 

It means that the whole array a^{→ d} can be represented as a sequence $$$a^{→ d} = [a_{n - d}, a_{n - d + 1}, …, a_{n - 1}, a_0, a_1, …, a_{n - d - 1}]$$$

  2. Then each element of the array a_i is replaced by a_i  \&  a^{→ d}_i, where \& is a logical "AND" operator. 



For example, if a = [0, 0, 1, 1] and d = 1, then a^{→ d} = [1, 0, 0, 1] and the value of a after the first step will be [0  \&  1, 0  \&  0, 1  \&  0, 1  \&  1], that is [0, 0, 0, 1].

The process ends when the array stops changing. For a given array a, determine whether it will consist of only zeros at the end of the process. If yes, also find the number of steps the process will take before it finishes.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases. 

The first line of each test case description contains two integers: n (1 ≤ n ≤ 10^6) — array size and d (1 ≤ d ≤ n) — cyclic shift offset. The second line of the description contains n space-separated integers a_i (0 ≤ a_i ≤ 1) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^6.

Output

Print t lines, each line containing the answer to the corresponding test case. The answer to a test case should be a single integer — the number of steps after which the array will contain only zeros for the first time. If there are still elements equal to 1 in the array after the end of the process, print -1.

Example

Input


5
2 1
0 1
3 2
0 1 0
5 2
1 1 0 1 0
4 2
0 1 0 1
1 1
0


Output


1
1
3
-1
0

Note

In the third sample test case the array will change as follows: 

  1. At the beginning a = [1, 1, 0, 1, 0], and a^{→ 2} = [1, 0, 1, 1, 0]. Their element-by-element "AND" is equal to $$$[1  \&  1, 1  \&  0, 0  \&  1, 1  \&  1, 0  \&  0] = [1, 0, 0, 1, 0]$$$ 
  2. Now a = [1, 0, 0, 1, 0], then a^{→ 2} = [1, 0, 1, 0, 0]. Their element-by-element "AND" equals to $$$[1  \&  1, 0  \&  0, 0  \&  1, 1  \&  0, 0  \&  0] = [1, 0, 0, 0, 0]$$$ 
  3. And finally, when a = [1, 0, 0, 0, 0] we get a^{→ 2} = [0, 0, 1, 0, 0]. Their element-by-element "AND" equals to $$$[1  \&  0, 0  \&  0, 0  \&  1, 0  \&  0, 0  \&  0] = [0, 0, 0, 0, 0]$$$ 

Thus, the answer is 3 steps.

In the fourth sample test case, the array will not change as it shifts by 2 to the right, so each element will be calculated as 0  \&  0 or 1  \&  1 thus not changing its value. So the answer is -1, the array will never contain only zeros.

import java.util.*;
import java.util.concurrent.CompletableFuture;

import javax.swing.text.Segment;


import java.io.*;
import java.math.*;
import java.sql.Array;



public class Main {
	  static class Reader{
		  
			BufferedReader br;
			StringTokenizer st;
			public Reader() {
				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()); 
		    } 

		    String nextLine() 
		    { 
		        String str = ""; 
		        try
		        { 
		            str = br.readLine(); 
		        } 
		        catch (IOException e) 
		        { 
		            e.printStackTrace(); 
		        } 
		        return str; 
		    } 
		}
	 
	  static long mod = (long)(1e9 + 7);
	 
	static void sort(long[] arr ) {
		 ArrayList<Long> al = new ArrayList<>();
		 for(long e:arr) al.add(e);
		 Collections.sort(al);
		 for(int i = 0 ; i<al.size(); i++) arr[i] = al.get(i);
	 }
	static void sort(int[] arr ) {
		 ArrayList<Integer> al = new ArrayList<>();
		 for(int e:arr) al.add(e);
		 Collections.sort(al);
		 for(int i = 0 ; i<al.size(); i++) arr[i] = al.get(i);
	 }
	static void sort(char[] arr) {
		ArrayList<Character> al = new ArrayList<Character>();
		for(char cc:arr) al.add(cc);
		Collections.sort(al);
		for(int i = 0 ;i<arr.length ;i++) arr[i] = al.get(i);
	}
	static void rvrs(int[] arr) {
		int i =0 , j = arr.length-1;
		while(i>=j) {
			int temp = arr[i];
			arr[i] = arr[j];
			arr[j] = temp;
			i++;
			j--;
		}
	}
	static void rvrs(long[] arr) {
		int i =0 , j = arr.length-1;
		while(i>=j) {
			long temp = arr[i];
			arr[i] = arr[j];
			arr[j] = temp;
			i++;
			j--;
		}
	}

	static long mod_mul(long a , long b ,long mod) {
		return ((a%mod)*(b%mod))%mod;
	}
	 static long gcd(long a, long b)
	  {      
	     if (b == 0)
	        return a;
	     return gcd(b, a % b); 
	  }
	 static boolean[] prime(int num) {
			boolean[] bool = new boolean[num];
		     
		      for (int i = 0; i< bool.length; i++) {
		         bool[i] = true;
		      }
		      for (int i = 2; i< Math.sqrt(num); i++) {
		         if(bool[i] == true) {
		            for(int j = (i*i); j<num; j = j+i) {
		               bool[j] = false;
		            }
		         }
		      }
		      if(num >= 0) {
		    	  bool[0] = false;
		    	  bool[1] = false;
		      }
		      
		      return bool;
		}

	 static long modInverse(long a, long m)
	    {
	        long g = gcd(a, m);
	       
	          return   power(a, m - 2, m);
	        
	    }
	   
	 static long power(long x, long y, long m){
	        if (y == 0) return 1; long p = power(x, y / 2, m) % m; p = (int)((p * (long)p) % m);
	        if (y % 2 == 0) return p; else return (int)((x * (long)p) % m); }
	   
     static class Combinations{
	    	 private long[] z;
	    	 private long[] z1;
	    	 private long[] z2;
	    	  public Combinations(long N , long mod) {
				z = new long[(int)N+1];
				z1 = new long[(int)N+1];
				z[0] = 1;
				for(int i =1 ; i<=N ; i++) z[i] = (z[i-1]*i)%mod;
			     z2 = new long[(int)N+1];
				z2[0] = z2[1] = 1;
			    for (int i = 2; i <= N; i++)
			        z2[i] = z2[(int)(mod % i)] * (mod - mod / i) % mod;
			    
			    
			    z1[0] = z1[1] = 1;
			    
			    for (int i = 2; i <= N; i++)
			        z1[i] = (z2[i] * z1[i - 1]) % mod;
			}
	    	  long fac(long n) {
	    		  return z[(int)n];
	    	  }
	    	  long invrsNum(long n) {
	    		  return z2[(int)n];
	    	  }
	    	  long invrsFac(long n) {
	    		  return invrsFac((int)n);
	    	  }
	    	  long ncr(long N, long R, long mod)
	    	  {		if(R<0 || R>N ) return 0;
	    		    long ans = ((z[(int)N] * z1[(int)R])
	    		              % mod * z1[(int)(N - R)])
	    		             % mod;
	    		    return ans;
	    		}
	      }
	      static class DisjointUnionSets {
			    int[] rank, parent;
			    int n;
			  
			    public DisjointUnionSets(int n)
			    {
			        rank = new int[n];
			        parent = new int[n];
			        this.n = n;
			        makeSet();
			    }
			  
			    void makeSet()
			    {
			        for (int i = 0; i < n; i++) {
			          
			            parent[i] = i;
			        }
			    }
			  
			    int find(int x)
			    {
			        if (parent[x] != x) {
			        
			            parent[x] = find(parent[x]);
			  
			        }
			  
			        return parent[x];
			    }
			  
			    void union(int x, int y)
			    {
			        int xRoot = find(x), yRoot = find(y);
			  
			        if (xRoot == yRoot)
			            return;
			  
			        if (rank[xRoot] < rank[yRoot])
			  
			            parent[xRoot] = yRoot;
			  
			        else if (rank[yRoot] < rank[xRoot])
			  
			            parent[yRoot] = xRoot;
			  
			        else
			        {
			            parent[yRoot] = xRoot;
			  
			            rank[xRoot] = rank[xRoot] + 1;
			        }
			    }
			}
	      static int[] KMP(String str) {
	    	  int n = str.length();
	    	  int[] kmp = new int[n];
	    	  for(int i = 1 ; i<n ; i++) {
	    		  int j = kmp[i-1];
	    		  while(j>0 && str.charAt(i) != str.charAt(j)) {
	    			  j = kmp[j-1];
	    		  }
	    		  if(str.charAt(i) == str.charAt(j)) j++;
	    		  kmp[i] = j;
	    	  }
	    	  
	    	  return kmp;
	      }
	      
	      
/************************************************ Query **************************************************************************************/	  
	 
/***************************************** 		Sparse Table	********************************************************/
	      static class SparseTable{
	    		
	    		private long[][] st;
	    		
	    		SparseTable(long[] arr){
	    			int n = arr.length;
	    			st = new long[n][25];
	    			log = new int[n+2];
	    			build_log(n+1);
	    			build(arr);
	    		}
	    		
	    		private void build(long[] arr) {
	    			int n = arr.length;
	    			
	    			for(int i = n-1 ; i>=0 ; i--) {
	    				for(int j = 0 ; j<25 ; j++) {
	    					int r = i + (1<<j)-1;
	    					if(r>=n) break;
	    					if(j == 0 ) st[i][j] = arr[i];
	    					else st[i][j] = Math.min(st[i][j-1] , st[ i + ( 1 << (j-1) ) ][ j-1 ] );
	    				}
	    			}
	    		}
	    		public long gcd(long a  ,long b) {
	    			if(a == 0) return b;
	    			return gcd(b%a , a);
	    		}
	    		public long query(int l ,int r) {
	    			int w = r-l+1;
	    			int power = log[w];
	    			return Math.min(st[l][power],st[r - (1<<power) + 1][power]);
	    		}
	    		private int[] log;
	    		void build_log(int n) {
	    			log[1] = 0;
	    			for(int i = 2 ; i<=n ; i++) {
	    				log[i] = 1 + log[i/2];
	    			}
	    		}
	    	}
	      
	      
/********************************************************	Segement Tree	*****************************************************/
/**
	 	 static class SegmentTree{
			 long[] tree;
			 long[] arr;
			 int n;
			 SegmentTree(long[] arr){
				 this.n = arr.length;
				 tree = new long[4*n+1];
				 this.arr = arr;
				 buildTree(0, n-1, 1);
			 }
			 
			 
			  void buildTree(int s ,int e  ,int index ) {
					if(s == e) {
						tree[index] = arr[s];
						return;
					}
				
					int mid = (s+e)/2;
					
					buildTree( s,  mid, 2*index);
					buildTree( mid+1, e, 2*index+1);
					
					tree[index] = Math.min(tree[2*index] , tree[2*index+1]);
				}
			  
			 long query(int si ,int ei) {
				 return query(0 ,n-1 , si ,ei , 1   );
			 }
			 private long query( int ss ,int se ,int qs , int qe,int index) {
					
					if(ss>=qs && se<=qe) return tree[index];
					
					if(qe<ss || se<qs) return (long)(1e17);
					
					int mid = (ss + se)/2;
					long left = query( ss , mid , qs ,qe , 2*index);
					long right= query(mid + 1 , se , qs ,qe , 2*index+1);
					return Math.min(left, right);
				}
			 public void update(int index , int val) {
				 arr[index] = val;
				 for(long e:arr) System.out.print(e+" ");
				 update(index , 0 , n-1 , 1);
			 }
			 private void update(int id ,int si , int ei , int index) {
				 if(id < si || id>ei) return;
				 if(si == ei ) { 
					 tree[index] = arr[id];
					 return;
				 }
				 if(si > ei) return;
				 int mid = (ei + si)/2;
					
					update( id,  si, mid , 2*index);
					update( id , mid+1, ei , 2*index+1);
					
					tree[index] = Math.min(tree[2*index] ,tree[2*index+1]);
			 }
			  
		 }
		 */

/* ***************************************************************************************************************************************************/	 
	 
	      static Reader sc = new Reader();
	 static StringBuilder sb = new StringBuilder();
	 public static void main(String args[]) throws IOException {
	
		 int tc = 1;
		  tc = sc.nextInt();

		 for(int i = 1 ; i<=tc ; i++) {

//			 sb.append("Case #" + i + ": "  );	// During KickStart && HackerCup
			 TEST_CASE();
			 
		 }
		 System.out.println(sb);
//		 System.out.println();
	 }

	static void TEST_CASE() {
		int n = sc.nextInt() , d = sc.nextInt();
		int[] arr = new int[n];
		for(int i = 0 ; i<n ; i++) {
			arr[i] = sc.nextInt();
		}
		long[] time = new long[n];
		Arrays.fill(time, -1);
		boolean[] visit = new boolean[n];
		for(int i =0 ; i<n ; i++) {
			if(visit[i] || arr[i] == 1) continue;
			
			dfs(i, d, arr, time, visit,0 );
		}
		long ans = 0 ;
		for(int i = 0 ; i<n ; i++) {
			if(arr[i] == 1 && time[i] == -1) {
				sb.append("-1\n");
				return;
			}
			ans = Math.max(ans, time[i]);
		}
		sb.append(ans+"\n");
	}
	static void dfs(int u , int d ,int[] arr, long[] time ,boolean[] visit,long val ) {
		if(visit[u]) return;
		visit[u] = true;
		if(arr[u] == 1) {
			time[u] = val+1;
		}else {
			arr[u] = 0;
		}
		if(arr[u] == 1) {
			dfs((u+d)%arr.length, d, arr, time, visit, val+1);
		}else {
			dfs((u+d)%arr.length, d, arr, time, visit, 0);
		}
	}
}



/*******************************************************************************************************************************************************/

/**

 */






