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 MAXN = (1 << 20);
int a[MAXN], x;
int posB[MAXN], N;
int it[22][MAXN], LOW, UP;
void scan() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", &x);
    posB[x] = i + 1;
  }
  for (int i = 1; i <= N; ++i) {
    scanf("%d", a + i);
    a[i] = posB[a[i]];
  }
}
void go(int l, int r, int lvl) {
  if (l == r) {
    it[lvl][l] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  go(l, mid, lvl + 1);
  go(mid + 1, r, lvl + 1);
  merge(it[lvl + 1] + l, it[lvl + 1] + mid + 1, it[lvl + 1] + mid + 1,
        it[lvl + 1] + r + 1, it[lvl] + l);
}
void makeTree() { go(1, N, 0); }
int query(int l, int r, int lvl, int L, int R) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R) {
    int* lw = lower_bound(it[lvl] + l, it[lvl] + r + 1, LOW);
    int* up = upper_bound(it[lvl] + l, it[lvl] + r + 1, UP);
    return up - lw;
  }
  int mid = (l + r) / 2;
  return query(l, mid, lvl + 1, L, R) + query(mid + 1, r, lvl + 1, L, R);
}
int query(int l1, int r1, int l2, int r2) {
  LOW = l1;
  UP = r1;
  return query(1, N, 0, l2, r2);
}
void solve() {
  makeTree();
  int M;
  scanf("%d", &M);
  x = 0;
  while (M--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = (a - 1 + x) % N + 1;
    b = (b - 1 + x) % N + 1;
    if (a > b) swap(a, b);
    c = (c - 1 + x) % N + 1;
    d = (d - 1 + x) % N + 1;
    if (c > d) swap(d, c);
    printf("%d\n", x = query(a, b, c, d));
    ++x;
  }
}
int main() {
  scan();
  solve();
}