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 <stdio.h>
#include <unordered_map>

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

std::unordered_map<int, int> mapQ;

int mina(int a, int b){
	return a < b ? a : b;
}

int maxa(int a, int b){
	return a > b ? a : b;
}

const int queryLength = 4;

int main(int argc, char * argv){
	int n = 0;
	scanf("%i", &n);

	int *p, q = 0;
	p = new int[n];

	int i = 0, x = 0, j = 0;
	while(i < n){
		scanf("%i", &p[i]);
		++i;
	}

	i = 0;
	while(i < n){
		scanf("%i", &q);
		mapQ[q] = ++i;
	}	

	int queryCount = 0;
	scanf("%i", &queryCount);
	int **query = new int*[queryCount];
	i = 0;
	while(i < queryCount){
		query[i] = new int[queryLength];
		j = 0;
		while(j < queryLength) { scanf("%i", &query[i][j]); ++j; }
		++i;
	}

	while(j < queryCount){

		int answer = 0;
		int a = f(query[j][0], x, n), b = f(query[j][1], x, n), c = f(query[j][2], x, n), d = f(query[j][3], x, n);

		int l1 = mina(a, b),
			r1 = maxa(a, b),
			l2 = mina(c, d),
			r2 = maxa(c, d);
		
		i = 0;
		while(i < n){
			int indP = i + 1, indQ = mapQ[p[i]];
			if((indP >= l1 && indP <= r1) && (indQ >= l2 && indQ <= r2)){
				++answer;
			}
			++i;
		}

		printf("%i\n", answer);
		
		x = answer + 1;	   
		++j;
	}

	//delete[] p,q, query;
	return 0;
}