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<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

const int Maxn=1000010;

int n,m;

struct Btree{
	int lx,rx;
	int sum;
};
Btree tree[Maxn*20];
int nume;

void insert(int &x,int y,int left,int right,int pos){
	x=++nume;
	
	tree[x].lx=tree[y].lx;
	tree[x].rx=tree[y].rx;
	tree[x].sum=tree[y].sum;
	
	tree[x].sum++;
	if (left==right){
		
	}else{
		int mid=(left+right)/2;
		if (pos<=mid) insert(tree[x].lx,tree[y].lx,left,mid,pos);
		if (mid<pos) insert(tree[x].rx,tree[y].rx,mid+1,right,pos);
	}
}

int query(int x,int left,int right,int l,int r){
	if (!x) return 0;
	if (l<=left && right<=r){
		return tree[x].sum;
	}else{
		int mid=(left+right)/2;
		int ret=0;
		if (l<=mid) ret+=query(tree[x].lx,left,mid,l,r);
		if (mid<r) ret+=query(tree[x].rx,mid+1,right,l,r);
		return ret;
	}
}

int Root[Maxn];


struct Node{
	int pos;
	int num;
};
Node node[Maxn];
int tmp;
int x=0;

inline bool cmpNum(Node a,Node b){
	if (a.pos<b.pos) return true;
	return false;
}

inline int fx(int num,int x){
	return (num-1+x)%n+1;
}

int main(){
	scanf("%d",&n);
	
	for (int i=1;i<=n;i++){
		scanf("%d",&tmp);
		node[tmp].pos=i;
	}
	for (int i=1;i<=n;i++){
		scanf("%d",&tmp);
		node[tmp].num=i;
	}
	
	sort(node+1,node+n+1,cmpNum);
	
	Root[0]=0;
	for (int i=1;i<=n;i++){
		insert(Root[i],Root[i-1],1,n,node[i].num);
	}
	
	scanf("%d",&m);
	for (int i=1;i<=m;i++){
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		a=fx(a,x);b=fx(b,x);c=fx(c,x);d=fx(d,x);
		int l1,r1,l2,r2;
		l1=min(a,b);r1=max(a,b);
		l2=min(c,d);r2=max(c,d);
		//printf("Query %d %d %d %d\n",l1,r1,l2,r2);
		printf("%d\n",query(Root[r1],1,n,l2,r2)-query(Root[l1-1],1,n,l2,r2));
		x++;
	}
	return 0;
}
