#include <iostream>

#include <string.h>

#include <cmath>

using namespace std;

const int N = 1e5 + ;

int n,k;

int s[N<<],sum[N<<];

int Q[N<<];

int main ()

{

    freopen("in.txt","r",stdin);

    int T; scanf("%d",&T);

    while (T--) {

        memset(Q,,sizeof(Q));

        memset(sum,,sizeof(sum));

        scanf("%d %d", &n, &k);

        for(int i=; i<=n; i++)

            scanf("%d",&s[i]), s[i+n]=s[i];

        n<<=;

        for(int i=;i<=n;i++)

            sum[i] = sum[i-]+s[i]; //printf("%d ",sum[i]);

        // for(int i=1;i<=n;i++)

        //     printf("%d ", sum[i]);

        // puts("");

        int mx = -N, ans1=n+,ans2=-;

        for(int i=;i<=n;i++) {

            if(i <= k) {

                if(sum[i] > mx) {

                    mx = sum[i];

                    ans1=,ans2=i;

                }

            }

        }

        //printf("%d %d\n",ans1, ans2);

        int st=,ed=;

        for(int i=; i<=n; i++)    {

            //[i-k+1,i] 区间长度为k

            while (st < ed && Q[st] < i-k)

                st++;

            if(st < ed) {

                int ans = sum[i] - sum[Q[st]];

                if(mx < ans) {

                    mx =ans;

                    ans1 = Q[st]+;

                    ans2 = i;

                }

            }

            while (st < ed && sum[i] <= sum[Q[ed-]])

                ed--;

            Q[ed++] = i;

            //cout << i<<" "<<ans1 <<" "<<ans2<<endl;

        }

        printf("%d %d %d\n", mx, ans1>n/?ans1-n/:ans1, ans2>n/?ans2-n/:ans2);

    }

    return ;

}