B - K-periodic Garland

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这篇博客介绍了一个关于字符串处理的算法问题——如何找到使 Garland 成为给定周期的最小操作次数。作者通过给出的 C++ 代码展示了如何使用动态规划策略来求解这个问题，其中涉及字符串遍历、状态转移和最优化策略。文章适用于对算法和动态规划感兴趣的读者。

链接：https://vjudge.net/contest/399019#problem/B

You are given a garland consisting of nn lamps. States of the lamps are represented by the string ss of length nn. The ii-th character of the string sisi equals '0' if the ii-th lamp is turned off or '1' if the ii-th lamp is turned on. You are also given a positive integer kk.

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 kk-periodic if the distance between each pair of adjacent turned on lamps is exactly kk. Consider the case k=3k=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 kk-periodic garland from the given one.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤25 0001≤t≤25 000) — the number of test cases. Then tt test cases follow.

The first line of the test case contains two integers nn and kk (1≤n≤106;1≤k≤n1≤n≤106;1≤k≤n) — the length of ss and the required period. The second line of the test case contains the string ss consisting of nn characters '0' and '1'.

It is guaranteed that the sum of nn over all test cases does not exceed 106106 (∑n≤106∑n≤106).

Output

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

Example

Input

Output

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn=1e6+7;
const ll mod=1e9+7;
ll t,n,k,s,ans;
char x[maxn];
ll dp[maxn][2];
ll a[maxn];
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        cin>>x;
        s=0;
        ans=mod;
        for(int i=0;i<n;i++)
        {
            if(x[i]=='1')
                s++;
            dp[i][0]=0;
            dp[i][1]=0;
            a[i]=s;
        }
            if(x[0]=='1')
                dp[0][1]=0,dp[0][0]=1;
            else
                dp[0][0]=0,dp[0][1]=1;
            ans=min(ans,dp[0][1]+a[n-1]-a[0]);
            ans=min(ans,dp[0][0]+a[n-1]-a[0]);
            for(int i=1;i<n;i++)
            {
                if(x[i]=='1')
                dp[i][1]=0,dp[i][0]=1;
                else
                dp[i][0]=0,dp[i][1]=1;
                dp[i][0]+=min(dp[i-1][0],dp[i-1][1]);
                if(i-k>=0)
                {
                    dp[i][1]+=min(dp[i-k][1]+a[i-1]-a[i-k],a[i-1]);
                }
                else
                {
                    dp[i][1]+=a[i-1];
                }
                ans=min(ans,dp[i][1]+a[n-1]-a[i]);
                ans=min(ans,dp[i][0]+a[n-1]-a[i]);
            }
            cout<<ans<<endl;
    }
    return 0;
}