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>

#define rep(i,_l,_r) for(register signed i=(_l),_end=(_r);i<=_end;++i)
#define fep(i,_l,_r) for(register signed i=(_l),_end=(_r);i>=_end;--i)
#define erep(i,u) for(signed i=head[u],v=to[i];i;i=nxt[i],v=to[i])
#define efep(i,u) for(signed i=Head[u],v=to[i];i;i=nxt[i],v=to[i])
#define print(x,y) write(x),putchar(y)

template <class T> inline T read(const T sample) {
    T x=0; int f=1; char s;
    while((s=getchar())>'9'||s<'0') if(s=='-') f=-1;
    while(s>='0'&&s<='9') x=(x<<1)+(x<<3)+(s^48),s=getchar();
    return x*f;
}
template <class T> inline void write(const T x) {
    if(x<0) return (void) (putchar('-'),write(-x));
    if(x>9) write(x/10);
    putchar(x%10^48);
}
template <class T> inline T Max(const T x,const T y) {if(x>y) return x; return y;}
template <class T> inline T Min(const T x,const T y) {if(x<y) return x; return y;}
template <class T> inline T fab(const T x) {return x>0?x:-x;}
template <class T> inline T gcd(const T x,const T y) {return y?gcd(y,x%y):x;}
template <class T> inline T lcm(const T x,const T y) {return x/gcd(x,y)*y;}

#include <iostream>
using namespace std;

const int maxn=1e6+5;

int n,ans,m,cnt,pos[maxn],b[maxn],rt[maxn];
struct Tree {
	int son[2],siz;
} t[maxn*20];

int f(int x) {return (x-1+ans)%n+1;}

void Insert(int pre,int &o,int l,int r,int k) {
	o=++cnt;
	t[o]=t[pre]; ++t[o].siz;
	if(l==r) return;
	int mid=l+r>>1;
	if(k<=mid) Insert(t[pre].son[0],t[o].son[0],l,mid,k);
	else Insert(t[pre].son[1],t[o].son[1],mid+1,r,k);
}

int Query(int o,int l,int r,int L,int R) {
	if(l>R||r<L) return 0;
	if(l>=L&&r<=R) return t[o].siz;
	int mid=l+r>>1;
	return Query(t[o].son[0],l,mid,L,R)+Query(t[o].son[1],mid+1,r,L,R);
}

int main() {
	n=read(9);
	rep(i,1,n) pos[read(9)]=i;
	rep(i,1,n) b[i]=pos[read(9)],Insert(rt[i-1],rt[i],1,n,b[i]);
	m=read(9);
	int a,b,c,d;
	while(m--) {
		a=f(read(9)),b=f(read(9)),c=f(read(9)),d=f(read(9));
		if(a>b) swap(a,b);
		if(c>d) swap(c,d);
		print(ans=Query(rt[d],1,n,a,b)-Query(rt[c-1],1,n,a,b),'\n');
		++ans;
	}
	return 0;
}