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;

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

#define N 1000010

int a[N], b[N];

struct node{
	int lc, rc, sz;
}t[N * 25];
int rt[N], TOT, n, m;

vector <int> g[N];

#define ls t[x].lc
#define rs t[x].rc
void modify(int &x, int p, int v, int l, int r){
	t[x = ++ TOT] = t[p];
	t[x].sz ++;
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(v <= mid) modify(ls, t[p].lc, v, l, mid);
	else modify(rs, t[p].rc, v, mid + 1, r);
}

int ask(int ql, int qr, int l, int r, int x){
	if(ql > qr || !x) return 0;
	if(ql <= l && r <= qr) return t[x].sz;
	int mid = (l + r) >> 1, ret = 0;
	if(ql <= mid) ret += ask(ql, qr, l, mid, ls);
	if(mid < qr)  ret += ask(ql, qr, mid + 1, r, rs);
	return ret;
}

int main(){
	scanf("%d", &n);
	int l, r, x, y, la = 0;
	rep(i, 1, n) scanf("%d", &x), a[x] = i;
	rep(i, 1, n) scanf("%d", &x), b[x] = i;
	rep(i, 1, n) g[a[i]].push_back(b[i]);
	rep(i, 1, n){
		rt[i] = rt[i - 1];
		for(int j = 0;j < g[i].size();j ++) modify(rt[i], rt[i], g[i][j], 1, n);
	}
	scanf("%d", &m);
	rep(i, 1, m){
		scanf("%d%d%d%d", &l, &r, &x, &y);
		l = (l - 1 + la) % n + 1;
		r = (r - 1 + la) % n + 1;
		x = (x - 1 + la) % n + 1;
		y = (y - 1 + la) % n + 1;
		if(l > r) swap(l, r);
		if(x > y) swap(x, y);
		la = ask(1, r, 1, n, rt[y]) + ask(1, l - 1, 1, n, rt[x - 1]) - ask(1, r, 1, n, rt[x - 1]) - ask(1, l - 1, 1, n, rt[y]);
		printf("%d\n", la);
		la ++;
	}
	return 0;
}

		
	
	
