You are given two permutations p and q, consisting of n elements, and m queries of the form: l1, r1, l2, r2 (l1 ≤ r1; l2 ≤ r2). The response for the query is the number of such integers from 1 to n, that their position in the first permutation is in segment [l1, r1] (borders included), and position in the second permutation is in segment [l2, r2] (borders included too).

A permutation of n elements is the sequence of n distinct integers, each not less than 1 and not greater than n.

Position of number v (1 ≤ v ≤ n) in permutation g1, g2, ..., gn is such number i, that gi = v.

Input

The first line contains one integer n (1 ≤ n ≤ 106), the number of elements in both permutations. The following line contains n integers, separated with spaces: p1, p2, ..., pn (1 ≤ pi ≤ n). These are elements of the first permutation. The next line contains the second permutation q1, q2, ..., qn in same format.

The following line contains an integer m (1 ≤ m ≤ 2·105), that is the number of queries.

The following m lines contain descriptions of queries one in a line. The description of the i-th query consists of four integers: a, b, c, d (1 ≤ a, b, c, d ≤ n). Query parameters l1, r1, l2, r2 are obtained from the numbers a, b, c, d using the following algorithm: 

  1. Introduce variable x. If it is the first query, then the variable equals 0, else it equals the response for the previous query plus one. 
  2. Introduce function f(z) = ((z - 1 + x) mod n) + 1. 
  3. Suppose l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)), r2 = max(f(c), f(d)). 

Output

Print a response for each query in a separate line.

Examples

Input

3
3 1 2
3 2 1
1
1 2 3 3


Output

1


Input

4
4 3 2 1
2 3 4 1
3
1 2 3 4
1 3 2 1
1 4 2 3


Output

1
1
2
1
1
1
1
1 1 1 1

1

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define Nil NULL
const int SIZEN=1000010;
class Node{
public:
	int l,r;
	int sum;
	Node *lc,*rc;
	Node* modify(int x,int t){
		if(l>x||r<x) return this;
		Node *p=new Node;
		*p=*this;
		p->sum+=t;
		if(p->lc!=Nil) p->lc=p->lc->modify(x,t);
		if(p->rc!=Nil) p->rc=p->rc->modify(x,t);
		return p;
	}
	int query(int a,int b){
		if(l>b||r<a) return 0;
		if(l>=a&&r<=b) return sum;
		return lc->query(a,b)+rc->query(a,b);
	}
};
Node* build(int a,int b){
	Node *p=new Node;
	p->l=a,p->r=b;
	p->sum=0;
	if(a==b) p->lc=p->rc=Nil;
	else{
		int mid=(a+b)/2;
		p->lc=build(a,mid);
		p->rc=build(mid+1,b);
	}
	return p;
}
Node T[SIZEN][20];
int N,M;
int P[SIZEN],Q[SIZEN];
int posQ[SIZEN];
Node *root[SIZEN];
int lastans=0;
int f(int z){
	return (z-1+lastans)%N+1;
}
void work(void){
	root[0]=build(1,N);
	for(int i=1;i<=N;i++)
		root[i]=root[i-1]->modify(posQ[P[i]],1);
	scanf("%d",&M);
	int a,b,c,d;
	for(int i=1;i<=M;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		a=f(a),b=f(b),c=f(c),d=f(d);
		if(a>b) swap(a,b);
		if(c>d) swap(c,d);
		lastans=root[b]->query(c,d)-root[a-1]->query(c,d);
		printf("%d\n",lastans);
		lastans++;
	}
}
void read(void){
	scanf("%d",&N);
	for(int i=1;i<=N;i++) scanf("%d",&P[i]);
	for(int i=1;i<=N;i++) scanf("%d",&Q[i]);
	for(int i=1;i<=N;i++) posQ[Q[i]]=i;
}
int main(){
	//freopen("t1.in","r",stdin);
	//read();
	//work();
	return 0;
}
