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<bits/stdc++.h>
using namespace std;
const int MAXN=1000010;
struct Tree{
	int lson,rson;
	int sum;
}t[MAXN*40];

int n;
int a[MAXN];
int q;
int tot;
int Map[MAXN];
int last;
int root[MAXN];

int update(int d,int x,int l,int r,int k){
	k=++tot;
	t[k]=t[d];
	t[k].sum=t[d].sum+1;
	if(l==r){
		return tot;
	}
	int mid=l+r>>1;
	if(x<=mid){
		t[k].lson=update(t[d].lson,x,l,mid,t[k].lson);
	}else{
		t[k].rson=update(t[d].rson,x,mid+1,r,t[k].rson);
	}
	return k;
}

int query(int l,int r,int tl,int tr,int k){
	if(l<=tl&&tr<=r){
		return t[k].sum;
	}
	int mid=tl+tr>>1,ans=0;
	if(l<=mid){
		ans+=query(l,r,tl,mid,t[k].lson);
	}
	if(r>mid){
		ans+=query(l,r,mid+1,tr,t[k].rson);
	}
	return ans;
}

int f(int k){
	return (k-1+last)%n+1;
}

int main(){
	ios::sync_with_stdio(false);
	scanf("%d",&n);
	int x;
	for(int i=1;i<=n;i++){
		scanf("%d",&x);
		Map[x]=i;
	}
	for(int i=1;i<=n;i++){
		scanf("%d",&x);
		a[i]=Map[x];
	}
	for(int i=1;i<=n;i++){
		root[i]=update(root[i-1],a[i],1,n,root[i]);
	}
	scanf("%d",&q);
	while(q--){
		int a,b,c,d;
		cin>>a>>b>>c>>d;
		int l1=min(f(a),f(b)),r1=max(f(a),f(b)),l2=min(f(c),f(d)),r2=max(f(c),f(d));
		//cout<<l1<<" "<<r1<<" "<<l2<<" "<<r2<<endl;
		last=query(l1,r1,1,n,root[r2])-query(l1,r1,1,n,root[l2-1]);
		printf("%d\n",last);
		last++;
	}
//	for(int i=1;i<=tot;i++){
//		cout<<t[i].lson<<" "<<t[i].rson<<endl;
//	}
	return 0;
}
