// you're already the best
// _
// ^ ^ //
// >(O_O)<___//
// \ __ __ \
// \\ \\ \\\\
#include "homecoming.h"
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3","unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using ll = long long;
//#define int long long
//#define double long double
#define forn(i,n) for(int i=0; i<(n); ++i)
#define pb push_back
#define pi pair<int,int>
#define f first
#define s second
#define vii(a,n) vector<int> a(n); forn(i,n) cin>>a[i];
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int inf = 1e18;
const int mod = 998244353;
void shift(vector<ll>&a) {
int t=a[0];
forn(i,a.size()-1) a[i]=a[i+1];
a[a.size()-1]=t;
}
long long solve(int N, int K, int *A, int *B) {
#define int long long
const int inf = 1e18;
int ans=0;
int n=N, k=K;
vector<int> a,b;
forn(i,n) a.pb(A[i]); forn(i,n) a.pb(a[i]);
forn(i,n) b.pb(B[i]); forn(i,n) b.pb(b[i]); forn(i,n) b.pb(b[i]);
vector<int> pra(2*n+1,0), prb(3*n+1,0);
forn(i,2*n) pra[i+1]=pra[i]+a[i];
forn(i,3*n) prb[i+1]=prb[i]+b[i];
vector<int> dp(2*n+1,0);
for (int i=1; i<=2*n; ++i) {
dp[i]=max(dp[i],dp[i-1]);
int sub=(i>=n)?dp[i-n]:0;
for (int j=max(0ll,i-n); j<i; ++j) {
dp[i]=max(dp[i],dp[j]-sub + (pra[i]-pra[j]) - (prb[i+k-1]-prb[max(j,i+k-1-n)]) );
}
}
return dp[2*n];
#undef int
}
