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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<int,int>
#define FOR(I,A,B) for(int I = int(A); I < int(B); ++I)

const int MN = 1000000;

P tree[MN];//tree[頂点番号]={子左,子右};
P value[MN];//value[頂点番号]={xの値,yの値};
int ind=1,root=1;

int n;
auto fx = [](P &a,P &b){return (a.first<b.first);};//x小さい順
auto fy = [](P &a,P &b){return (a.second<b.second);};//y小さい順

void make_tree(int p,vector<P> data,int depth,bool leftc){
	sort(data.begin(), data.end(),(depth%2?fy:fx));//0x 1y
	int mid = data.size();mid/=2;
	if(leftc){
		tree[p].first = ind;
		value[ind] = data[mid];
	}else{
		tree[p].second = ind;
		value[ind] = data[mid];
	}
	int pp = ind;
	ind++;
	vector<P> datal,datar;
	FOR(i,0,mid)datal.push_back(data[i]);
	FOR(i,mid+1,data.size())datar.push_back(data[i]);
	data.clear();
	if(datal.size())make_tree(pp,datal,depth+1,true);
	if(datar.size())make_tree(pp,datar,depth+1,false);
}

vector<P> ans;
int sx,tx,sy,ty;
void calc(int ind,int dep){
	int x = value[ind].first;
	int y = value[ind].second;
	if(sx<=x&&x<=tx && sy<=y&&y<=ty){
		ans.push_back({x,y});
		if(~tree[ind].first)calc(tree[ind].first,dep+1);
		if(~tree[ind].second)calc(tree[ind].second,dep+1);
		return;
	}
	if(dep&1){//y
		if(~tree[ind].first){
			if(value[ind].second>=sy)calc(tree[ind].first,dep+1);
		}
		if(~tree[ind].second){
			if(value[ind].second<=ty)calc(tree[ind].second,dep+1);
		}
	}else{//x
		if(~tree[ind].first){//-1じゃない
			if(value[ind].first>=sx)calc(tree[ind].first,dep+1);
		}
		if(~tree[ind].second){
			if(value[ind].first<=tx)calc(tree[ind].second,dep+1);
		}
	}
}


int main(){

	FOR(i,0,MN)tree[i]=value[i]={-1,-1};
	cin >> n;
	vector<P> xy(n);
	FOR(i,0,n)cin>>xy[i].first>>xy[i].second;
	map<P,int> ma;
	FOR(i,0,n)ma[xy[i]] = i;
	make_tree(0,xy,0,true);
	/*FOR(i,0,7){
		cout << "i:" << i << " " << tree[i].first << " " << tree[i].second << " : " << value[i].first << " " << value[i].second << endl;
	}*/
	int q;
	cin >> q;
	FOR(i,0,q){
		ans.clear();
		cin >> sx >> tx >> sy >> ty;
		if(n)calc(1,0);
		vector<int> ansi;
		for(P x:ans)ansi.push_back(ma[x]);
		sort(ansi.begin(), ansi.end());
		for(int i:ansi) cout << i << '\n';
		cout << '\n';
	}
}
