Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INPUT_MAXCHARS = 1 << 16;
char buffer[INPUT_MAXCHARS];
struct FastReader {
  char *p;
  FastReader() : p(buffer) { fread(buffer, 1, sizeof buffer, stdin); }
  int next_int() {
    int remchars = INPUT_MAXCHARS - (p - buffer);
    if (remchars < 25) {
      memcpy(buffer, p, remchars);
      size_t cnt = fread(buffer + remchars, 1, sizeof buffer - remchars, stdin);
      if (remchars + cnt < sizeof buffer) {
        buffer[remchars + cnt] = 0;
      }
      p = buffer;
    }
    while (*p < '0') {
      p++;
    }
    int val = 0;
    while (*p >= '0') {
      val = val * 10 + (*p - '0');
      ++p;
    }
    return val;
  }
};
const int MAXPOW = 1 << 20;
vector<int> T[2 * MAXPOW];
const int MAXN = 1000000;
int revP[MAXN], revQ[MAXN];
int query(int x, int a, int b, int l1, int r1, int l2, int r2) {
  if (a >= r1 || b <= l1) return 0;
  if (l1 <= a && b <= r1) {
    return lower_bound(T[x].begin(), T[x].end(), r2) -
           lower_bound(T[x].begin(), T[x].end(), l2);
  } else {
    return query(2 * x, a, (a + b) / 2, l1, r1, l2, r2) +
           query(2 * x + 1, (a + b) / 2, b, l1, r1, l2, r2);
  }
}
int main() {
  FastReader reader;
  int n = reader.next_int();
  for (int i = 0; i < n; ++i) {
    int x = reader.next_int();
    revP[x - 1] = i;
  }
  for (int i = 0; i < n; ++i) {
    int x = reader.next_int();
    revQ[x - 1] = i;
  }
  int sz = 1;
  while (sz < n) {
    sz *= 2;
  }
  for (int i = 0; i < n; ++i) {
    T[sz + revP[i]].push_back(revQ[i]);
  }
  for (int i = sz - 1; i > 0; --i) {
    int left = 2 * i;
    int right = left + 1;
    merge(T[left].begin(), T[left].end(), T[right].begin(), T[right].end(),
          back_inserter(T[i]));
  }
  int Q = reader.next_int();
  int x = 0;
  while (Q--) {
    int a = reader.next_int();
    int b = reader.next_int();
    int c = reader.next_int();
    int d = reader.next_int();
    int l1 = min((a - 1 + x) % n, (b - 1 + x) % n);
    int r1 = max((a - 1 + x) % n, (b - 1 + x) % n);
    int l2 = min((c - 1 + x) % n, (d - 1 + x) % n);
    int r2 = max((c - 1 + x) % n, (d - 1 + x) % n);
    x = query(1, 0, sz, l1, r1 + 1, l2, r2 + 1);
    cout << x << '\n';
    ++x;
  }
  return 0;
}