The range search problem consists of a set of attributed records S to determine which records from S intersect with a given range.

For n points on a plane, report a set of points which are within in a given range. Note that you do not need to consider insert and delete operations for the set.

Constraints

* 0 ≤ n ≤ 500,000
* 0 ≤ q ≤ 20,000
* -1,000,000,000 ≤ x, y, sx, tx, sy, ty ≤ 1,000,000,000
* sx ≤ tx
* sy ≤ ty
* For each query, the number of points which are within the range is less than or equal to 100.

Input


n
x0 y0
x1 y1
:
xn-1 yn-1
q
sx0 tx0 sy0 ty0
sx1 tx1 sy1 ty1
:
sxq-1 txq-1 syq-1 tyq-1


The first integer n is the number of points. In the following n lines, the coordinate of the i-th point is given by two integers xi and yi.

The next integer q is the number of queries. In the following q lines, each query is given by four integers, sxi, txi, syi, tyi.

Output

For each query, report IDs of points such that sxi ≤ x ≤ txi and syi ≤ y ≤ tyi. The IDs should be reported in ascending order. Print an ID in a line, and print a blank line at the end of output for the each query.

Example

Input

6
2 1
2 2
4 2
6 2
3 3
5 4
2
2 4 0 4
4 10 2 5


Output

0
1
2
4

2
3
5
#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define NMAX (500000)
#define NIL (-1)

typedef struct {
	int id;
	int x;
	int y;
} Point;

typedef struct {
	int pid;
	int ppos;
	int left;
	int right;
} Node;

Point P[NMAX];
Node T[NMAX];
int num = 0;
priority_queue<int, vector<int>, greater<int> > ans;

bool lessX(Point& p1, Point& p2) { return p1.x < p2.x; }
bool lessY(Point& p1, Point& p2) { return p1.y < p2.y; }

void setPoint(int n)
{
	int x, y;

	rep(i, n) {
		scanf("%d %d", &x, &y);
		P[i].id = i;
		P[i].x = x;
		P[i].y = y;
	}
}

void init_kDTree(int n)
{
	rep(i, n) {
		T[i].pid = NIL;
		T[i].ppos = NIL;
		T[i].left = NIL;
		T[i].right = NIL;
	}
}

int make_kDTree(int left, int right, int depth)
{
	int mid, index;

	if (left >= right) return NIL;

	if (depth % 2 == 0) sort(P + left, P + right, lessX);
	else sort(P + left, P + right, lessY);

	mid = (left + right) / 2;
	index = num++;

	T[index].pid = P[mid].id;
	T[index].ppos = mid;
	T[index].left = make_kDTree(left, mid, depth + 1);
	T[index].right = make_kDTree(mid + 1, right, depth + 1);

	return index;
}

void search_kDTree(int index, int depth, Point* sp, Point* tp)
{
	int Pid, Ppos, Px, Py, Tleft, Tright;

	Pid = T[index].pid;
	Ppos = T[index].ppos;
	Tleft = T[index].left;
	Tright = T[index].right;

	Px = P[Ppos].x;
	Py = P[Ppos].y;

	if (((sp->x) <= Px && Px <= (tp->x))
		&& ((sp->y) <= Py && Py <= (tp->y))) {
		ans.push(Pid);
	}

	if (depth % 2 == 0) {
		if ((sp->x) <= Px && Tleft != NIL) search_kDTree(Tleft, depth + 1, sp, tp);
		if ((tp->x) >= Px && Tright != NIL) search_kDTree(Tright, depth + 1, sp, tp);
	}
	else {
		if ((sp->y) <= Py && Tleft != NIL) search_kDTree(Tleft, depth + 1, sp, tp);
		if ((tp->y) >= Py && Tright != NIL) search_kDTree(Tright, depth + 1, sp, tp);
	}
}

int main()
{
	int n, q, sx, tx, sy, ty;
	Point sp, tp;

	scanf("%d", &n);

	setPoint(n);
	init_kDTree(n);
	make_kDTree(0, n, 0);

	scanf("%d", &q);
	rep(i, q) {
		scanf("%d %d %d %d", &sx, &tx, &sy, &ty);
		sp.x = sx;
		sp.y = sy;
		tp.x = tx;
		tp.y = ty;

		search_kDTree(0, 0, &sp, &tp);

		while (!ans.empty()) {
			printf("%d\n", ans.top());
			ans.pop();
		}
		putchar('\n');
	}

	return 0;
}
