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 = 100010;
int arr[maxn];
int pos[maxn];
int N, M;
int dest[maxn];
struct SegNode {
  SegNode *L, *R;
  int sum;
};
SegNode nodes[maxn * 25];
int C;
struct SegTree {
  SegNode *null;
  void init() {
    C = 0;
    null = &nodes[C++];
    null->L = null->R = null;
    null->sum = 0;
  }
  SegNode *update(int pos, int l, int r, SegNode *root, int value) {
    SegNode *x = &nodes[C++];
    x->L = root->L;
    x->R = root->R;
    x->sum = root->sum;
    if (l == r) {
      x->sum += value;
      return x;
    }
    int mid = (l + r) / 2;
    if (pos <= mid)
      x->L = update(pos, l, mid, root->L, value);
    else
      x->R = update(pos, mid + 1, r, root->R, value);
    x->sum = x->L->sum + x->R->sum;
    return x;
  }
  int query(int left, int right, int l, int r, SegNode *root) {
    if (left == l && right == r) {
      return root->sum;
    }
    int mid = (l + r) / 2;
    if (right <= mid) {
      return query(left, right, l, mid, root->L);
    } else if (left > mid) {
      return query(left, right, mid + 1, r, root->R);
    } else {
      return query(left, mid, l, mid, root->L) +
             query(mid + 1, right, mid + 1, r, root->R);
    }
  }
};
SegTree tree;
SegNode *root[maxn];
int ar[maxn];
void init() {
  tree.init();
  root[0] = tree.null;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", arr + i);
  }
  for (int i = 1; i <= N; i++) {
    scanf("%d", ar + i);
    pos[ar[i]] = i;
  }
  for (int i = 1; i <= N; i++) {
    root[i] = tree.update(pos[i], 0, N, root[i - 1], 1);
  }
}
void work() {
  int last = 0;
  scanf("%d", &M);
  for (int i = 1; i <= M; i++) {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    a = (a + last - 1) % N + 1;
    b = (b + last - 1) % N + 1;
    c = (c + last - 1) % N + 1;
    d = (d + last - 1) % N + 1;
    if (a > b) swap(a, b);
    if (c > d) swap(c, d);
    last =
        tree.query(c, d, 0, N, root[b]) - tree.query(c, d, 0, N, root[a - 1]);
    printf("%d\n", last);
    last++;
  }
}
int main() {
  init();
  work();
  return 0;
}
