//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
//#pragma GCC target("bmi,bmi2,popcnt,lzcnt")
#include "homecoming.h"
#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cmath>
#include <fstream>
#include <climits>
#include <queue>
#include <algorithm>
#include <stdint.h>
#include <stack>
#include <iomanip>
#include <unordered_set>
#include <unordered_map>
#include <cstring> // Для memset
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair <ll, ll> pii;
typedef pair <ld, ld> pdd;
const ll DIM = 700007;
const ll MXMASK = (1 << 21);
const ll INF = 1e18;
const ll mod = 998244353;
const ld eps = 0.00000000001;
const ld gr = (sqrt(5) + 1) / 2;
const ll offset = 10000;
const ll Bits = 20;
const ll dx[4] = { 1, 0, -1, 0 };
const ll dy[4] = { 0, 1, 0, -1 };
FILE* stream;
long long int solve(int N, int K, int* A, int* B) {
vector < ll > p(2 * N + 1, 0);
long long int res = 0;
p[0] = B[0];
for (int i = 1; i < 2 * N; i++) {
if (i < N) p[i] = p[i - 1] + B[i];
else p[i] = p[i - 1] + B[i - N];
}
if(A[0] - p[K] >= 0) res = A[0] - p[K - 1];
for (int i = 1; i < N; i++) {
if (-p[i + K - 1] + p[i - 1] + A[i] >= 0) res += -p[i + K - 1] + p[i - 1] + A[i];
}
return res;
}
/*
3 2
40 80 100
140 0 20
*/
