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 <cstring>
#include <algorithm>
using namespace std;

const int N = int(1e6 + 6);
const int Nlg = 21;

struct rc
{
	int cnt;
	rc *l, *r;
}nil[N*Nlg], *root[N], *tot;

int a[N], b[N], c[N];
int n, m, x;

rc* ins(rc *x, int l, int r, int p)
{
	rc *ret = ++tot;
	if(x!=nil) (*ret) = (*x); else ret->l=ret->r=nil;
	++ret->cnt;
	
	if(l==r) return ret;
	
	int mid((l+r)>>1);
	if(p<=mid) ret->l = ins(ret->l, l, mid, p);
	else ret->r = ins(ret->r, mid+1, r, p);
	
	return ret;
}

int Q(rc *x, int l, int r, int p, int q)
{
	if(l==p && r==q) return x->cnt;
	if(x==nil) return 0;
	
	int mid((l+r)>>1);
	if(q<=mid) return Q(x->l, l, mid, p, q);
	else if(p>mid) return Q(x->r, mid+1, r, p, q);
	return Q(x->l, l, mid, p, mid) + Q(x->r, mid+1, r, mid+1, q);
}

int r()
{
	int ret(0), c; while(c=getchar(), c<'0');
	ret = c - '0'; while(c=getchar(), c>='0') ret = ret * 10 + c - '0';
	return ret;
}

void T(int &a){a = (a-1+x+1) % n + 1;}

int main()
{
	
	n = r();
	for(int i=1; i<=n; i++) a[i] = r(), b[a[i]] = i;
	for(int i=1; i<=n; i++) c[i] = r(), c[i] = b[c[i]];
	
	root[0] = tot = nil;
	for(int i=1; i<=n; i++) root[i] = ins(root[i-1], 1, n, c[i]);
	
	m = r(), x = -1;
	for(int u, v, p, q; m; --m)
	{
		u = r(), v = r(), p = r(), q = r();
		T(u), T(v); if(u>v) swap(u, v);
		T(p), T(q); if(p>q) swap(p, q);
		printf("%d\n", x = Q(root[v], 1, n, p, q) - Q(root[u-1], 1, n, p, q));
	}
	
	return 0;
}

