根据题意我们知道，不能选两个有交集的区间，我们只需要先选一遍最大的长度为k的区间，然后再遍历一遍选择大小为第二大的区间。

选择第二大的区间我的方案是不与上一个选出来的产生交集

// Problem: 数学考试
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/problem/15553
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// Code by: ING__
// 
// Powered by CP Editor (https://cpeditor.org)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#define ll long long
#define re return
#define Endl "\n"
#define endl "\n"

using namespace std;

typedef pair<int, int> PII;

int dx[4] = {-1,0,1,0};
int dy[4] = {0,1,0,-1};

int T;
int n;
int k;
ll s[200010];

int main(){
    cin >> T;
    while(T--){
        cin >> n >> k;

        memset(s, 0, sizeof(s));

        for(int i = 1; i <= n; i++){
            ll x;
            cin >> x;
            s[i] = s[i - 1] + x;
        }

        // int ans = 0;

        // 选取第一大的区间

        ll maxa = -1e19;
        int r1;

        for(int i = k; i <= n; i++){
             if(maxa < s[i] - s[i - k]){
                 maxa = s[i] - s[i - k];
                 r1 = i;
             }
        }

        // 选取第二大的区间

        ll maxb = -1e19;

        for(int i = k; i <= n; i++){
            if(i >= r1 - k + 1 && i - k + 1 <= r1) continue;
            if(maxb < s[i] - s[i - k]){
                maxb = s[i] - s[i - k];
            }
        }

        cout << maxa + maxb << endl;
    }
    re 0;
}