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<algorithm>
#include<vector>
using namespace std;

struct tree{
	int x,y;
	int num;
}tree[500000];
vector<int>ans;
bool cmpx(struct tree a , struct tree b){
	return a.x < b.x;
}

bool cmpy(struct tree a, struct tree b){
	return a.y < b.y;
}

void build(int begin,int end,int level){
	if(begin<end){
		if(!level)
			sort(tree+begin,tree+end,cmpx);
		else
			sort(tree+begin,tree+end,cmpy);
		int mid=(end+begin)/2;
		build(begin,mid,level^1);
		build(mid+1,end,level^1);
	}
}

void search(int begin ,int end, int sx , int tx ,int sy ,int ty ,int level ){
	if(begin<end){
		int mid=(end+begin)/2;
		if(!level){
			if(tree[mid].x>=sx&&tree[mid].x<=tx){
				if(tree[mid].y>=sy&&tree[mid].y<=ty)
					ans.push_back(tree[mid].num);
				search(begin,mid,sx,tx,sy,ty,level^1);
				search(mid+1,end,sx,tx,sy,ty,level^1);
			}
			else if(tx<tree[mid].x) search(begin,mid,sx,tx,sy,ty,level^1);
			else if(sx>tree[mid].x)	search(mid+1,end,sx,tx,sy,ty,level^1);
		}
		else{
			if(tree[mid].y>=sy&&tree[mid].y<=ty){
				if(tree[mid].x>=sx&&tree[mid].x<=tx)
					ans.push_back(tree[mid].num);
				search(begin,mid,sx,tx,sy,ty,level^1);
				search(mid+1,end,sx,tx,sy,ty,level^1);
			}
			else if(ty<tree[mid].y) search(begin,mid,sx,tx,sy,ty,level^1);
			else if(sy>tree[mid].y)	search(mid+1,end,sx,tx,sy,ty,level^1);
		}
	}
}

int main(){
	int n,p,x,y,sx,tx,sy,ty;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d %d",&x,&y);
		tree[i].x=x;
		tree[i].y=y;
		tree[i].num=i;
	}
	build(0,n,0);
	scanf("%d",&p);
	while(p--){
		scanf("%d %d %d %d",&sx,&tx,&sy,&ty);
		ans.clear();
		search(0,n,sx,tx,sy,ty,0);
		sort(ans.begin(),ans.end());
		for(int i=0;i<ans.size();i++)printf("%d\n",ans[i]);
		printf("\n");
	}
	
}
