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<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

class Node {
public:
  int location;
  int parent;
  int right;
  int left;

  Node() {}
};

class Point {
public:
  int id, x, y;
  Point() {}
  Point(int id, int x, int y)
    : id(id)
    , x(x)
    , y(y)
    {}

  bool operator < (const Point& p) const {
    return id < p.id;
  }

  void print() {
    printf("%d\n", id);
  }
};

static const int MAX = 1000000;
static const int NIL = -1;

int N;
Point P[MAX];
Node T[MAX];

int np;

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

int makeKDTree(int left, int right, int depth) {
  if (!(left < right)) {
    return NIL;
  }
  int mid = (left + right) / 2;
  int t = np++;
  if (depth % 2 == 0) {
    sort(P + left, P + right, lessX);
  } else {
    sort(P + left, P + right, lessY);
  }
  T[t].location = mid;
  T[t].left = makeKDTree(left, mid, depth + 1);
  T[t].right = makeKDTree(mid + 1, right, depth + 1);

  return t;
}

void find(int v, int sx, int tx, int sy, int ty, int depth, vector<Point>& ans) {
  int x = P[T[v].location].x;
  int y = P[T[v].location].y;

  if (sx <= x && x <= tx && sy <= y && y <= ty) {
    ans.push_back(P[T[v].location]);
  }

  if (depth % 2 == 0) {
    if (T[v].left != NIL) {
      if (sx <= x) {
        find(T[v].left, sx, tx, sy, ty, depth + 1, ans);
      }
    }
    if (T[v].right != NIL) {
      if (x <= tx) {
        find(T[v].right, sx, tx, sy, ty, depth + 1, ans);
      }
    }
  } else {
    if (T[v].left != NIL) {
      if (sy <= y) {
        find(T[v].left, sx, tx, sy, ty, depth + 1, ans);
      }
    }
    if (T[v].right != NIL) {
      if (y <= ty) {
        find(T[v].right, sx, tx, sy, ty, depth + 1, ans);
      }
    }
  }
}

int main() {
  int x, y;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d %d", &x, &y);
    P[i] = Point(i, x, y);
    T[i].left = T[i].right = T[i].parent = NIL;
  }

  np = 0;

  int root = makeKDTree(0, N, 0);

  int q;
  scanf("%d", &q);
  int sx, tx, sy, ty;
  vector<Point> ans;
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d %d", &sx, &tx, &sy, &ty);
    ans.clear();
    find(root, sx, tx, sy, ty, 0, ans);
    sort(ans.begin(), ans.end());
    for (int j = 0; j < ans.size(); j++) {
      ans[j].print();
    }
    printf("\n");
  }

  return 0;
}


