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<algorithm>
#include<cstdio>

using namespace std;

#define rep(i,a,b) for (int i=(a);i<=(b);i++)

struct tree{
	int ls,rs,sum;
	tree() {ls=rs=sum=0;}
};

const int NN=1000005;

int n,m,v,cnt,a[NN],b[NN],rt[NN];
tree t[30*NN];

int read()
{
	int x=0;char ch=getchar();
	while (ch<'0'||ch>'9') ch=getchar();
	while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x;
}
void Put(int &ME,int YOU,int L,int R,int x)
{
	ME=++cnt;t[ME]=t[YOU];t[ME].sum++;
	if (L==R) return;int mid=(L+R)>>1;
	if (x<=mid) Put(t[ME].ls,t[YOU].ls,L,mid,x);
	else Put(t[ME].rs,t[YOU].rs,mid+1,R,x);
}
int Get(int id,int L,int R,int l,int r)
{
	if (!id||!t[id].sum) return 0;
	if (L==l&&R==r) return t[id].sum;
	int mid=(L+R)>>1;
	if (r<=mid) return Get(t[id].ls,L,mid,l,r);
	else if (l>mid) return Get(t[id].rs,mid+1,R,l,r);
	else return Get(t[id].ls,L,mid,l,mid)+Get(t[id].rs,mid+1,R,mid+1,r);
}
int f(int z) {return (z-1+v)%n+1;}
int main()
{
	n=read();rep(i,1,n) a[i]=read();rep(i,1,n) b[read()]=i;
	rep(i,1,n) Put(rt[i],rt[i-1],1,n,b[a[i]]);
	for (v=0,m=read();m;m--,v++)
	{
		int A=read(),B=read(),C=read(),D=read();
		int l1=min(f(A),f(B)),r1=max(f(A),f(B));
		int l2=min(f(C),f(D)),r2=max(f(C),f(D));
		printf("%d\n",Get(rt[r1],1,n,l2,r2)-Get(rt[l1-1],1,n,l2,r2));
	}
	return 0;
}