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>
namespace mystd {
    inline int read() {
        int w = 1, q = 0;
        char ch = ' ';
        while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
        if (ch == '-') w = -1, ch = getchar();
        while (ch >= '0' && ch <= '9') q = q * 10 + ch - '0', ch = getchar();
        return w * q;
    }
    inline void write(int x) {
        if (x < 0) {
            x = ~(x - 1);
            putchar('-');
        }
        if (x > 9) write(x / 10);
        putchar(x % 10 + '0');
    }
}
using namespace std;
using namespace mystd;

const int maxn = 2e6 + 200;
struct tree { int ls, rs, sum; } tr[maxn << 4];
int n, m, ans, cnt, s[maxn], rt[maxn], pos[maxn];

void update(int &cur, int l, int r, int x, int pre) {
    cur = ++cnt;
    tr[cur].ls = tr[pre].ls;
    tr[cur].rs = tr[pre].rs;
    tr[cur].sum = tr[pre].sum + 1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (x <= mid) update(tr[cur].ls, l, mid, x, tr[pre].ls);
    else update(tr[cur].rs, mid + 1, r, x, tr[pre].rs);
}
int query(int cur, int l, int r, int s, int t) {
    if (l >= s && r <= t) return tr[cur].sum;
    int mid = (l + r) >> 1, res = 0;
    if (s <= mid) res += query(tr[cur].ls, l, mid, s, t);
    if (t > mid) res += query(tr[cur].rs, mid + 1, r, s, t);
    return res;
}

int change(int x) { return (x - 1 + ans) % n + 1; }

int main() {
    n = read();
    for (int i = 1; i <= n; i++) {
        pos[read()] = i;
    }
    for (int i = 1; i <= n; i++) {
        s[i] = pos[read()];
    }
    for (int i = 1; i <= n; i++) {
        update(rt[i], 1, n, s[i], rt[i - 1]);
    }
    m = read();
    for (int i = 1, a, b, c, d; i <= m; i++) {
        a = read();
        b = read();
        c = read();
        d = read();
        if (a > b) swap(a, b);
        if (c > d) swap(c, d);
        ans = query(rt[d], 1, n, a, b) - query(rt[c - 1], 1, n, a, b);
        write(ans);
        puts("");
        ans++;
    }
    return 0;
}