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 dbg(x) cerr << #x << " = " << x << endl
#define rep(i, a, b) for(int i = (a); i <= (b); ++ i)

const int N = 1e6 + 10; 

template <typename Tp> void read(Tp &x){
	x = 0; int op = 1; char ch = getchar();
	while(!isdigit(ch)){ if(ch == '-') op = -1; ch = getchar(); }
	while(isdigit(ch)){ x = x*10+ch-'0'; ch = getchar(); } x *= op;
}
template <typename Tp> void CMax(Tp &x, Tp y){ if(y > x) x = y; }
template <typename Tp> void CMin(Tp &x, Tp y){ if(y < x) x = y; }

int n, m, p1[N], p2[N], q1[N], q2[N];
int rt[N], node;
struct segTree { int dat, ls, rs; } t[N * 40];

#define mid ((l + r) >> 1)

void modify(int &x, int l, int r, int pos) {
	t[++ node] = t[x]; x = node; t[x].dat ++;
	if(l == r) return;
	if(pos <= mid) modify(t[x].ls, l, mid, pos);
	else modify(t[x].rs, mid + 1, r, pos);
}
int Query(int x, int y, int l, int r, int L, int R) {
	if(L <= l && r <= R) return t[y].dat - t[x].dat;
	int sum = 0;
	if(L <= mid) sum += Query(t[x].ls, t[y].ls, l, mid, L, R);
	if(R > mid) sum += Query(t[x].rs, t[y].rs, mid + 1, r, L, R);
	return sum;
}

#undef mid

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

int main()
{
	read(n);
	rep(i, 1, n) read(p1[i]), p2[p1[i]] = i;
	rep(i, 1, n) read(q1[i]), q2[i] = p2[q1[i]], printf("%d\n", q2[i]);
	rep(i, 1, n) {
		rt[i] = rt[i - 1];
		modify(rt[i], 1, n, q2[i]);
	}
	read(m);
	int a, b, c, d, l1, r1, l2, r2;
	while(m -- ) {
		read(a); read(b); read(c); read(d);
		l1 = min(f(a), f(b)); r1 = max(f(a), f(b));
		l2 = min(f(c), f(d)); r2 = max(f(c), f(d));
		ans = Query(rt[l2 - 1], rt[r2], 1, n, l1, r1);
		printf("%d\n", ans); ans ++;
	}
	return 0;
}
