This is the easy version of the problem. The only difference is that in this version k = 0.

There is an array a_1, a_2, …, a_n of n positive integers. You should divide it into a minimal number of continuous segments, such that in each segment there are no two numbers (on different positions), whose product is a perfect square.

Moreover, it is allowed to do at most k such operations before the division: choose a number in the array and change its value to any positive integer. But in this version k = 0, so it is not important.

What is the minimum number of continuous segments you should use if you will make changes optimally?

Input

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

The first line of each test case contains two integers n, k (1 ≤ n ≤ 2 ⋅ 10^5, k = 0).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^7).

It's guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case print a single integer — the answer to the problem.

Example

Input


3
5 0
18 6 2 4 1
5 0
6 8 1 24 8
1 0
1


Output


3
2
1

Note

In the first test case the division may be as follows:

  * [18, 6] 
  * [2, 4] 
  * [1] 
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.List;
 import java.util.*;
  public class realfast implements Runnable 
  {
    private static final int INF = (int) 1e9;
    long in= 1000000007;
    long fac[]= new long[1000001];
    long inv[]=new long[1000001];
    public  void solve() throws IOException 
    {

         int t = readInt();
         boolean prime[]=new boolean[4000];

         Arrays.fill(prime,true);

         ArrayList<Integer> c= new ArrayList<Integer>();

         for(int i=2;i<4000;i++)
         {
            if(prime[i])
            {
                c.add(i);
                for(int j=2*i;j<4000;j=j+i)
                {
                    prime[j]= false;
                }
            }
         }
         int hash[]=new int[10000001];
         for(int f =0;f<t;f++)
         {
             int n = readInt();
             int k = readInt();
             int arr[]=new int[n];
             for(int i=0;i<n;i++)
                arr[i]= readInt();

             for(int i =0;i<n;i++)
             {
                for(int j =0;j<c.size();j++)
                {
                    int gal = (int)c.get(j);
                    gal = gal*gal;

                    while(arr[i]%gal==0)
                    {
                        arr[i]= arr[i]/gal;
                    }

                }

             }
             int start =0;
             for(int i=0;i<n;i++)
             {
                hash[arr[i]]=-1;
             }
             int count =1;
             for(int i =0;i<n;i++)
             {
                if(hash[arr[i]]>=start)
                {
                    count++;
                    start=i;
                }
                hash[arr[i]]= i;
             }
             out.println(count);

         }

        
      

    }
    
    public int value (int seg[], int left , int right ,int index, int l, int r)
    {
            
            if(left>right)
            {
              return -100000000;
            }
            if(right<l||left>r)
                return -100000000;
            if(left>=l&&right<=r)
                return seg[index];
            int mid = left+(right-left)/2;
            int val = value(seg,left,mid,2*index+1,l,r);
            int val2 = value(seg,mid+1,right,2*index+2,l,r);
            return Math.max(val,val2);

    }
   
    public int gcd(int a , int b )
    {
      if(a<b)
      {
        int t =a;
        a=b;
        b=t;
      }
      if(a%b==0)
        return b ;
      return gcd(b,a%b);
    }
    public long pow(long n , long p,long m)
    {
         if(p==0)
            return 1;
        long val = pow(n,p/2,m);;
        val= (val*val)%m;
        if(p%2==0)
            return val;
        else
            return (val*n)%m;
    }
    
    
    ///////////////////////////////////////////////////////////////////////////////////////////////////////////////////
    public static void main(String[] args) {
        new Thread(null, new realfast(), "", 128 * (1L << 20)).start();
    }
 
    private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    private BufferedReader reader;
    private StringTokenizer tokenizer;
    private PrintWriter out;
 
    @Override
    public void run() {
        try {
            if (ONLINE_JUDGE || !new File("input.txt").exists()) {
                reader = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            } else {
                reader = new BufferedReader(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");
            }
            solve();
        } catch (IOException e) {
            throw new RuntimeException(e);
        } finally {
            try {
                reader.close();
            } catch (IOException e) {
                // nothing
            }
            out.close();
        }
    }
 
    private String readString() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }
 
    @SuppressWarnings("unused")
    private int readInt() throws IOException {
        return Integer.parseInt(readString());
    }
 
    @SuppressWarnings("unused")
    private long readLong() throws IOException {
        return Long.parseLong(readString());
    }
 
    @SuppressWarnings("unused")
    private double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }
}
class edge implements Comparable<edge>{
    int u ;
    int v;
    
    edge(int u, int v)
    {
       this.u=u;
       this.v=v;
    }
    public int compareTo(edge e)
    {
        return this.v-e.v;
    }
}