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 BS_2 = 108;
const int BS = 216;
const int N = 1010101;
int n, m, a, b, c, d, a2, b2, c2, d2, p[N], q[N], p2[N], q2[N], f[4700][4700];
int ans;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", p + i);
    p2[--p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", q + i);
    q2[--q[i]] = i;
  }
  ans = n - 1;
  memset(f, 0, sizeof(f));
  for (int i = 0; i < n; i++) f[(i / BS) + 1][((p[i] = q2[p[i]]) / BS) + 1]++;
  for (int i = 1; i <= (n / BS) + 2; i++)
    for (int j = 1; j <= (n / BS) + 2; j++)
      f[i][j] += f[i][j - 1] + f[i - 1][j] - f[i - 1][j - 1];
  for (int i = 0; i < n; i++) q[p[i]] = i;
  for (int i = n; i <= n + 333; i++) p[i] = q[i] = -1;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = (a + ans) % n;
    b = (b + ans) % n;
    c = (c + ans) % n;
    d = (d + ans) % n;
    if (a > b) swap(a, b);
    if (c > d) swap(c, d);
    a2 = (a + BS_2) / BS;
    b2 = (b + BS_2) / BS;
    c2 = (c + BS_2) / BS;
    d2 = (d + BS_2) / BS;
    ans = f[b2][d2] - f[b2][c2] - f[a2][d2] + f[a2][c2];
    a2 *= BS;
    b2 *= BS;
    c2 *= BS;
    d2 *= BS;
    if (a2 < a) {
      for (int j = a2; j < a; j++)
        if (p[j] >= c && p[j] <= d) ans--;
    } else {
      for (int j = a; j < a2; j++)
        if (p[j] >= c && p[j] <= d) ans++;
    }
    if (b < b2) {
      for (int j = b + 1; j < b2; j++)
        if (p[j] >= c && p[j] <= d) ans--;
    } else {
      for (int j = b2; j <= b; j++)
        if (p[j] >= c && p[j] <= d) ans++;
    }
    if (c2 < c) {
      for (int j = c2; j < c; j++)
        if (q[j] >= a2 && q[j] < b2) ans--;
    } else {
      for (int j = c; j < c2; j++)
        if (q[j] >= a2 && q[j] < b2) ans++;
    }
    if (d < d2) {
      for (int j = d + 1; j < d2; j++)
        if (q[j] >= a2 && q[j] < b2) ans--;
    } else {
      for (int j = d2; j <= d; j++)
        if (q[j] >= a2 && q[j] < b2) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}