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 <set>
#include <map>
#include <cmath>
#include <stack>
#include <ctime>
#include <queue>
#include <cstdio>
#include <bitset>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read()
{
    int x = 0,f = 1;char c = getchar();
    while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
    while(isdigit(c)){x = x * 10+ c - '0',c=getchar();}
    return x*f;
}
const int maxn = 1e6 + 10;
int a[maxn], b[maxn];
struct SegmentTree
{
	int lc, rc, cnt;
}tree[maxn << 5];
int root[maxn], tot;
void build(int &p, int l, int r)
{
	p = ++tot;
	if(l == r)
	{
		return;
	}
	int mid = (l + r) >> 1;
	build(tree[p].lc, l, mid);
	build(tree[p].rc, mid + 1, r);
}
void insert(int now, int &p, int l, int r, int x)
{
	p = ++tot;
	tree[p] = tree[now];tree[p].cnt++;
	if(l == r)
	{
		return;
	}
	int mid = (l + r) >> 1;
	if(x <= mid)
	{
		insert(tree[now].lc, tree[p].lc, l, mid, x);
	}
	else
	{
		insert(tree[now].rc, tree[p].rc, mid + 1, r, x);
	}
}
int ask(int now, int l, int r, int x)
{
	if(x == 0)
	{
		return 0;
	}
	if(l == r)
	{
		return tree[now].cnt;
	}
	int mid = (l + r) >> 1;
	if(x <= mid)
	{
		return ask(tree[now].lc, l, mid, x);
	}
	else return ask(tree[now].rc, mid + 1, r, x) + tree[tree[now].lc].cnt;
}
int n, las;
int f(int x)
{
	return (x - 1 + las) % n + 1;
}
int main()
{
//	freopen("1.in", "r", stdin);
	n = read();build(root[0], 1, n);
	for (int i = 1; i <= n; ++i) a[i] = read();
	for (int i = 1; i <= n; ++i) b[read()] = i;
	for (int i = 1; i <= n; ++i) insert(root[i - 1], root[i], 1, n, b[a[i]]);
	int m = read();
	while (m--)
	{
		int a = read(), b = read(), c = read(), d = read();
		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));
//		ask(root[r1], 1, n, l2 - 1);
		printf("%d\n", (ask(root[r1], 1, n, r2) - ask(root[r1], 1, n, l2 - 1)) - (ask(root[l1 - 1], 1, n, r2) - ask(root[l1 - 1], 1, n, l2 - 1)));
	}
	return 0;
}

