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 = 1000000 + 3;
const int maxm = 200000 + 5;
int n;
int m;
int p1[maxn], p2[maxn], g2[maxn];
int value[maxn], name[maxn], par[maxn], lefT[maxn], righT[maxn];
vector<int> righTl[maxn];
int x;
void print() {
  for (int i = 0; i < n; i++) {
  }
}
void put(int l1, int l2) {
  int cur = 0;
  while (value[cur] != -1) {
    if (l2 > value[cur]) {
      righTl[cur].push_back(l1);
      if (righT[cur] != -1)
        cur = righT[cur];
      else {
        righT[cur] = l1;
        par[l1] = cur;
        cur = l1;
      }
    } else {
      if (lefT[cur] != -1)
        cur = lefT[cur];
      else {
        lefT[cur] = l1;
        par[l1] = cur;
        cur = l1;
      }
    }
  }
  value[cur] = l2;
  name[cur] = p1[l1];
}
int bs(vector<int> u, int l, int r, int key) {
  if (l >= r) return 0;
  if (r == l + 1) {
    if (u[l] == key) return l;
    return r;
  }
  int mid = (l + r) / 2;
  if (u[mid] > key) return bs(u, l, mid, key);
  return bs(u, mid, r, key);
}
int f(int a) { return ((a - 1 + x) % n); }
int find(int l1, int r1, int l, int root, bool aa) {
  if (root == -1) return 0;
  if (value[root] >= l) {
    int w = 0;
    if (root >= l1 && root <= r1 && (aa || value[root] != l)) w++;
    if (righTl[root].size() < 2) {
      if (righTl[root].size() == 1) {
        int y = righTl[root][0];
        if (y >= l1 && y <= r1) w++;
      }
    } else {
      int a = bs(righTl[root], 0, righTl[root].size() - 1, l1);
      int b = bs(righTl[root], 0, righTl[root].size() - 1, r1);
      if (righTl[root][b] == r1) w++;
      w += b - a;
    }
    if (value[root] == l) return w;
    root = lefT[root];
    return find(l1, r1, l, root, aa) + w;
  } else {
    root = righT[root];
    return find(l1, r1, l, root, aa);
  }
}
int query(int l1, int r1, int l2, int r2) {
  int a = find(l1, r1, l2, 0, 1);
  int b = find(l1, r1, r2, 0, 2);
  return a - b;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p1[i];
  for (int i = 0; i < n; i++) cin >> p2[i];
  for (int i = 0; i < n; i++) g2[p2[i]] = i;
  for (int i = 0; i < n; i++) {
    value[i] = -1;
    par[i] = -1;
    lefT[i] = -1;
    righT[i] = -1;
  }
  for (int i = 0; i < n; i++) put(i, g2[p1[i]]);
  cin >> m;
  int a, b, c, d;
  int l1, l2, r1, r2;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c >> 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));
    x = query(l1, r1, l2, r2);
    cout << x << endl;
    x++;
  }
  return 0;
}
