The problem can be solved in \(O(n)\) by the same algorithm of offcial solution, with Run-length Compression on the queue.

using pii = pair<int, int>;

const int MAX=200005;
int a[MAX];
int b[MAX];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int _t;
	cin>>_t;
	while(_t--){
		int n,d;
		cin>>n>>d;
		for(int i=0;i<n;i++){
			cin>>a[i];
		}
		for(int i=0;i<n;i++){
			cin>>b[i];
		}
		queue<pii> eggs;
		ll ans=0;
		for(int i=0;i<n;i++){
			eggs.emplace(i,a[i]);
			ans+=a[i];
			int t=b[i];
			ans-=b[i];
			while(t>0){
				auto& [j,cnt]=eggs.front();
				int s=min(t,cnt);
				t-=s;
				cnt-=s;
				if(s==0) eggs.pop();
			}
			while(!eggs.empty()){
				auto [j,cnt]=eggs.front();
				if(j>i-d) break;
				ans-=cnt;
				eggs.pop();
			}
		}
		cout<<ans<<'\n';
	}
}