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>
const int maxn = 1000000;
using namespace std;
int n, m, las, a[maxn], b[maxn], c[maxn];
vector<int> tr[maxn * 4];
void build(int p, int l, int r) {
  if (l == r) {
    tr[p].push_back(a[l]);
    return;
  }
  int mid = (l + r) / 2;
  build(p * 2, l, mid), build(p * 2 + 1, mid + 1, r);
  int ltop = 0, rtop = 0, llim = mid - l, rlim = r - mid - 1;
  for (int i = l; i <= r; ++i)
    if (rtop > rlim || (rtop <= rlim && ltop <= llim &&
                        tr[p * 2][ltop] <= tr[p * 2 + 1][rtop]))
      tr[p].push_back(tr[p * 2][ltop++]);
    else
      tr[p].push_back(tr[p * 2 + 1][rtop++]);
}
int f(int ave) { return ((ave - 1 + las) % n) + 1; }
int query(int p, int l, int r, int lx, int rx, int lans, int rans) {
  if (l == lx && r == rx) {
    int L = 0, R = r - l, ans = -1, ret = r - l + 1;
    while (L <= R) {
      int mid = (L + R) / 2;
      if (tr[p][mid] < lans)
        ans = mid, L = mid + 1;
      else
        R = mid - 1;
    }
    ret -= ans + 1;
    L = 0, R = r - l, ans = r - l + 1;
    while (L <= R) {
      int mid = (L + R) / 2;
      if (tr[p][mid] > rans)
        ans = mid, R = mid - 1;
      else
        L = mid + 1;
    }
    ret -= (r - l) - ans + 1;
    return ret;
  }
  int mid = (l + r) / 2;
  if (rx <= mid) return query(p * 2, l, mid, lx, rx, lans, rans);
  if (lx > mid) return query(p * 2 + 1, mid + 1, r, lx, rx, lans, rans);
  return query(p * 2, l, mid, lx, mid, lans, rans) +
         query(p * 2 + 1, mid + 1, r, mid + 1, rx, lans, rans);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]), c[b[i]] = i;
  for (int i = 1; i <= n; ++i) a[i] = c[a[i]];
  build(1, 1, n);
  scanf("%d", &m);
  las = 0;
  int A, B, C, D;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d%d", &A, &B, &C, &D);
    int l = min(f(A), f(B)), r = max(f(A), f(B)), lx = min(f(C), f(D)),
        rx = max(f(C), f(D));
    las = query(1, 1, n, l, r, lx, rx) + 1;
    printf("%d\n", las - 1);
  }
  return 0;
}