You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.

In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).

The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.

Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.

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

Output

For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.

Example

Input


6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0


Output


1
2
5
4
0
0
/*
    dp[i] is the minimum cost to fix the suffix till i. 
    pref[i] stores the number of 1's till i.
    ans[i] = minimum cost so that the first 1 is at i and its
             suffix is fixed.
             
    Suffix is fixed till i meaning: 
    For an i, we consider that the 1s are only at i%k, i%k+k, etc.
    So, if we keep str[i]==0, we need to set all characters after it
    to 0, else if we keep str[i]==1, cost to fix its suffix equals cost to 
    set all characters from i+1  to i+k-1 to 0, plus the cost to fix the 
    suffix at i+k.
 */

//created by Whiplash99
import java.io.*;
import java.util.*;
public class A
{
    public static void main(String[] args) throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N;

        int T=Integer.parseInt(br.readLine().trim());
        StringBuilder sb=new StringBuilder();

        while (T-->0)
        {
            String[] s=br.readLine().trim().split(" ");
            N=Integer.parseInt(s[0]);
            int K=Integer.parseInt(s[1]);

            char[] str=br.readLine().trim().toCharArray();
            int[] pref=new int[N+1];
            for(i=1;i<=N;i++) pref[i]=pref[i-1]+(str[i-1]-'0');

            int ans=pref[N];
            int[] dp=new int[N+1];

            for(i=N-1;i>=0;i--)
            {
                int next = Math.min(i+K,N);
                int nextCost=dp[next]+(pref[next]-pref[i+1]);

                if(str[i]=='1') dp[i]=nextCost;
                else dp[i]=Math.min(1+nextCost, pref[N]-pref[i+1]);
            }

            for(i=0;i<N;i++) ans=Math.min(ans,dp[i]+pref[i]);

            sb.append(ans).append("\n");
        }
        System.out.println(sb);
    }
}