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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int Maxn = 1000010, Maxm = 100005, Mo = 1000000007,
          sp[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long oo = INT_MAX >> 2;
struct Tree {
  int l, r, cnt;
};
struct edge {
  int u, v, len;
};
typedef int IArr[Maxn];
typedef long long LArr[Maxn];
typedef double DArr[Maxn];
IArr x, y, sx, sy;
int n, i, j, ans;
vector<int> vt[Maxn];
void ins(int pos, int key) {
  while (pos <= n) {
    vt[pos].push_back((key));
    pos += pos & -pos;
  }
}
int ask(int pos, int key) {
  int ans = 0;
  while (pos) {
    ans += upper_bound(vt[pos].begin(), vt[pos].end(), key) - vt[pos].begin();
    pos ^= pos & -pos;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    int a;
    cin >> a;
    x[a] = i;
  }
  for (i = 1; i <= n; i++) {
    int a;
    cin >> a;
    y[a] = i;
  }
  for (i = 1; i <= n; i++) ins(x[i], y[i]);
  for (i = 1; i <= n; i++) sort(vt[i].begin(), vt[i].end());
  int ct = 0, q, a, b, c, d;
  cin >> q;
  while (q--) {
    cin >> a >> b >> c >> d;
    a = (a - 1 + ct) % n + 1;
    b = (b - 1 + ct) % n + 1;
    c = (c - 1 + ct) % n + 1;
    d = (d - 1 + ct) % n + 1;
    int a1 = min(a, b), b1 = max(a, b), c1 = min(c, d), d1 = max(c, d);
    ct = ask(b1, d1) - ask(b1, c1 - 1) - ask(a1 - 1, d1) + ask(a1 - 1, c1 - 1);
    cout << ct++ << endl;
  }
}
