#include<bits/stdc++.h>
using namespace std;
const long long inf = 3000000000000000000;
vector<pair<int,int> >p[100005];
int n,q;
long long x[100005],y[100005];
bool vis[100005];
long long ans = -inf;
long long X,Y;vector<int>vv;
int fa[100005];
int find(int x){
	if(fa[x] == x)return x;return fa[x] = find(fa[x]);
}void merge(int x,int y){fa[find(x)] = find(y);}int id;
inline void dfs(int now){
	vv.push_back(now);vis[now] = 1;
	ans = max(ans,X*x[now]+Y*y[now]);
	for(int i =0;i<p[now].size();i++){
		if(vis[p[now][i].first] or p[now][i].second>=id)continue;
		dfs(p[now][i].first);
	}
}
signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >>n>>q;
	for(int i = 1;i<=n;i++)cin>>x[i];
	for(int i = 1;i<=n;i++)cin>>y[i];
	for(int i = 1;i<=n;i++)fa[i] = i;
	long long lastans = 0;
	for(int i = 1;i<=q;i++){
		int op;cin>>op;
		if(op == 0){
			long long u,v,s;cin>>u>>v>>s;
			u = (u+s*lastans)%n+1,v = (v+s*lastans)%n+1;
			if(find(u) == find(v))continue;merge(u,v);
			p[u].push_back({v,i}),p[v].push_back({u,i});
		}else{long long u,s;
			cin>>X>>Y>>u>>id>>s;
			u = (u+s*lastans)%n+1,id = (id+s*lastans)%i+1;
			vv.clear();ans = -inf;dfs(u);cout<<(lastans = ans)<<'\n';
			if(ans<0)lastans*=-1;
			for(int x:vv)vis[x] = 0;
		}
	}
	return 0;
}
/*
我肯定会 polylog 做法，能不能比 n2 快另说。
显然这个东西可以用李超树做，但是需要实现的东西是可持久化李超树合并加上可持久化并查集。
这我哪会啊？
*/