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<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define maxn 1000005
using namespace std;
struct node{
	node *lc,*rc;
	int s;
}nil[maxn*30],*rot[maxn],*cnt;
int a[maxn],n,ky,num[maxn];
int m,le,ri,k,preans=0,ple,pri;
char ch;

inline int add(int x,int y,const int ha){
	x+=y;
	if(x>=ha) return x-ha;
	else return x;
}

node *update(node *u,int l,int r){
	node *ret=++cnt;
	*ret=*u;
	ret->s++;
	
	if(l==r) return ret;
	
	int mid=l+r>>1;
	if(le<=mid) ret->lc=update(ret->lc,l,mid);
	else ret->rc=update(ret->rc,mid+1,r);
	
	return ret;
}

int query(node *u,node *v,int l,int r){
	if(l>=le&&r<=ri) return v->s-u->s;
	
	int mid=l+r>>1,an=0;
	if(le<=mid) an+=query(u->lc,v->lc,l,mid);
	if(ri>mid) an+=query(u->rc,v->rc,mid+1,r);
	return an;
}

inline void prework(){
	cnt=rot[0]=nil->lc=nil->rc=nil;
	nil->s=0;
	
	for(int i=1;i<=n;i++){
		le=a[i];
		rot[i]=update(rot[i-1],1,n);
	}
}

inline void solve(){
	scanf("%d",&m);
	while(m--){
		scanf("%d%d%d%d",&le,&ri,&ple,&pri);
		
		le=add(le,preans-1,n)+1;
		ri=add(ri,preans-1,n)+1;
		ple=add(ple,preans-1,n)+1;
		pri=add(pri,preans-1,n)+1;
		if(le>ri) swap(le,ri);
		if(ple>pri) swap(ple,pri);
				
		preans=query(rot[ple-1],rot[pri],1,n);
		printf("%d\n",preans);
	}
}

int main(){
	scanf("%d",&n);
	int now;
	for(int i=1;i<=n;i++){
		scanf("%d",&now);
		num[now]=i;
	}
	for(int i=1;i<=n;i++){
		scanf("%d",&now);
		a[i]=num[now];
	}
	
	prework();
	solve();
	
	return 0;
}
