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 N = 1e6 + 10, M = N * 10;
int n, m, a[N], pos[N], t;
int rt[N], lc[M], rc[M], cnt, sum[M];
inline void build(int &p, int l, int r) {
  p = ++cnt;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(lc[p], l, mid);
  build(rc[p], mid + 1, r);
}
inline void merg(int &p, int o, int l, int r, int x) {
  p = ++cnt;
  sum[p] = sum[o];
  sum[p]++;
  lc[p] = lc[o];
  rc[p] = rc[o];
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (x <= mid)
    merg(lc[p], lc[o], l, mid, x);
  else
    merg(rc[p], rc[o], mid + 1, r, x);
}
inline int query(int p, int o, int l, int r, int lim) {
  if (l > lim) return 0;
  if (r <= lim) return sum[p] - sum[o];
  int mid = (l + r) >> 1;
  int tmp = 0;
  if (mid <= lim) {
    tmp = sum[lc[p]] - sum[lc[o]] + query(rc[p], rc[o], mid + 1, r, lim);
  } else
    tmp = query(lc[p], lc[o], l, mid, lim);
  return tmp;
}
inline void solve(int l1, int r1, int l2, int r2) {
  int anss = query(rt[r2], rt[l2 - 1], 1, n, r1) -
             query(rt[r2], rt[l2 - 1], 1, n, l1 - 1);
  printf("%d\n", anss);
}
inline int fox(int x) { return (x - 1 + t) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    pos[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = pos[x];
  }
  build(rt[0], 1, n);
  for (int i = 1; i <= n; i++) merg(rt[i], rt[i - 1], 1, n, a[i]);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int l1, r1, l2, r2, x1, x2, x3, x4;
    scanf("%d%d%d%d", &x1, &x2, &x3, &x4);
    l1 = min(fox(x1), fox(x2)), r1 = max(fox(x1), fox(x2));
    l2 = min(fox(x3), fox(x4)), r2 = max(fox(x3), fox(x4));
    solve(l1, r1, l2, r2);
    t++;
  }
  return 0;
}
