// 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(n), b(n);
forn(i,n) a[i]=A[i], b[i]=B[i];
forn(i,n) b.pb(b[i]);
forn(it,n) {
vector<int> dp(n+1,0);
vector<int> pra(n+1,0), prb(2*n+1,0);
forn(i,n) pra[i+1]=pra[i]+a[i];
forn(i,2*n) prb[i+1]=prb[i]+b[i];
for (int i=1; i<=n; ++i) {
dp[i]=max(dp[i],dp[i-1]);
for (int j=0; j<i; ++j) {
//cout<<i<<' '<<j<<' '<<dp[j]<<' '<<pra[i]-pra[j]<<' '<<prb[i+k-1]-prb[max(j,i+k-1-n)]<<'\n';
dp[i]=max(dp[i],dp[j]+(pra[i]-pra[j])-(prb[i+k-1]-prb[max(j,i+k-1-n)]));
}
//cout<<dp[i]<<'\n';
}
//cout<<'\n';
ans=max(ans,dp[n]);
shift(a),shift(b);
}
return ans;
#undef int
}
