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
#include <bits/stdc++.h>
using namespace std;
inline void READ(int &x) {
  char c;
  x = 0;
  do c = getchar();
  while (c < '0' || c > '9');
  do x = x * 10 + c - 48, c = getchar();
  while (c >= '0' && c <= '9');
}
struct NODE {
  int v;
  NODE *l, *r;
  NODE() : v(0), l(0), r(0) {}
};
const int maxn = 1000005;
int N, M, pos[maxn];
NODE *root[maxn];
void Ins(const NODE *old, NODE *&x, int l, int r, int q) {
  x = new NODE();
  if (l == r) {
    x->v++;
    return;
  }
  int mid((l + r) >> 1);
  if (q <= mid)
    Ins(old ? old->l : 0, x->l, l, mid, q), x->r = (old ? old->r : 0);
  else
    Ins(old ? old->r : 0, x->r, mid + 1, r, q), x->l = (old ? old->l : 0);
  x->v = (x->l ? x->l->v : 0) + (x->r ? x->r->v : 0);
}
int Find(const NODE *x, int l, int r, int ql, int qr) {
  if (!x) return 0;
  if (l == ql && r == qr) return x->v;
  int mid((l + r) >> 1);
  if (qr <= mid) return Find(x->l, l, mid, ql, qr);
  if (ql > mid) return Find(x->r, mid + 1, r, ql, qr);
  return Find(x->l, l, mid, ql, mid) + Find(x->r, mid + 1, r, mid + 1, qr);
}
int main() {
  READ(N);
  for (int i = 1; i <= N; i++) {
    int x;
    READ(x);
    pos[x] = i;
  }
  for (int i = 1; i <= N; i++) {
    int x;
    READ(x);
    Ins(root[i - 1], root[i], 1, N, pos[x]);
  }
  READ(M);
  int ans = -1;
  for (int i = 1; i <= M; i++) {
    int a, b, c, d, l1, r1, l2, r2;
    READ(a), READ(b), READ(c), READ(d);
    l1 = min((((a + ans) % N) + 1), (((b + ans) % N) + 1)),
    r1 = max((((a + ans) % N) + 1), (((b + ans) % N) + 1)),
    l2 = min((((c + ans) % N) + 1), (((d + ans) % N) + 1)),
    r2 = max((((c + ans) % N) + 1), (((d + ans) % N) + 1));
    ans = Find(root[r2], 1, N, l1, r1) - Find(root[l2 - 1], 1, N, l1, r1);
    printf("%d\n", ans);
  }
  return 0;
}
