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;
const int maxn = 1000010;
struct node {
  int l, r;
  vector<int> info;
};
node tree[4 * maxn];
int n, m, l1, r1, l2, r2, x;
int a, b, c, d;
int p1[maxn], p2[maxn], ps[maxn], pos[maxn];
int f(int z) { return ((z - 1 + x) % n) + 1; }
void build(int rt, int l, int r) {
  tree[rt].info.clear();
  tree[rt].l = l;
  tree[rt].r = r;
  if (l == r) {
    tree[rt].info.push_back(pos[l]);
    return;
  }
  int mid = (l + r) >> 1;
  int i, p, q;
  build(2 * rt, l, mid);
  build(2 * rt + 1, mid + 1, r);
  for (i = 0, p = 0, q = 0; p <= mid - l + 1 && q <= r - mid; i++) {
    if (tree[2 * rt].info[p] < tree[2 * rt + 1].info[q])
      tree[rt].info.push_back(tree[2 * rt].info[p++]);
    else
      tree[rt].info.push_back(tree[2 * rt + 1].info[q++]);
  }
  if (p == mid - l + 1)
    for (; i < r - l + 1; i++)
      tree[rt].info.push_back(tree[2 * rt + 1].info[q++]);
  else
    for (; i < r - l + 1; i++) tree[rt].info.push_back(tree[2 * rt].info[p++]);
  return;
}
int query(int rt, int l, int r) {
  if (l <= tree[rt].l && tree[rt].r <= r) {
    int LL, RR;
    RR = upper_bound(tree[rt].info.begin(), tree[rt].info.end(), r2) -
         tree[rt].info.begin();
    LL = lower_bound(tree[rt].info.begin(), tree[rt].info.end(), l2) -
         tree[rt].info.begin();
    return RR - LL;
  }
  int res = 0, mid = (tree[rt].l + tree[rt].r) >> 1;
  if (l <= mid) res += query(2 * rt, l, r);
  if (mid < r) res += query(2 * rt + 1, l, r);
  return res;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &p1[i]);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &p2[i]);
      ps[p2[i]] = i;
    }
    for (int i = 1; i <= n; i++) pos[i] = ps[p1[i]];
    build(1, 1, n);
    scanf("%d", &m);
    x = 0;
    while (m--) {
      scanf("%d%d%d%d", &a, &b, &c, &d);
      l1 = f(a), r1 = f(b), l2 = f(c), r2 = f(d);
      if (l1 > r1) swap(l1, r1);
      if (l2 > r2) swap(l2, r2);
      printf("%d\n", x = query(1, l1, r1));
      x++;
    }
  }
  return 0;
}
