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>  /**!!!  注意此??数据一般?大 使用scanf printf?出?入*/
#include<algorithm>
#include<vector>//使用??数?存??果

using namespace std;

class Node{//?的?点
public:
    int location;//表示?点在数?中的索引
    int p,l,r;//
    Node(){}
};

class Point{
public:
    int id,x,y;//x,y?坐?
    Point(){}
    Point(int id,int x,int y):id(id),x(x),y(y){}
    bool operator < (const Point &p) const{
        return id<p.id;//用于按点的?入?序?出
    }
};

static const int MAX=1000000;//数?大小 用数???二叉搜索?
static const int NIL=-1;//定???

int N;//??点的数量
Point P[MAX];//存?点的集合
Node T[MAX];//??二叉搜索?的数?
int np;//生出??的索引??
int root;//存?根?点的索引
vector<Point> ans;//存?答案的数?

bool lessX(const Point&p1,const Point&p2){return p1.x<p2.x;}//按照x排序?使用
bool lessY(const Point&p1,const Point&p2){return p1.y<p2.y;}//按照y排序?使用

void init(){//初始化
    np=0;
    ans.clear();//清零
}
int makeKDTree(int l,int r,int depth){/**K代表几?的搜索  !!!不包括r*/
    if(!(l<r))
        return NIL;
    int mid=(l+r)/2;//?次都从中点?始
    int t=np++;//??移?
    if(depth%2==0){//?数?偶数?
        sort(P+l,P+r,lessX);//按照x升序排序 不包括r
    }
    else{
        sort(P+l,P+r,lessY);//当?数?奇数? 按照y升序排序 不包括r
    }
    T[t].location=mid;//??当前?点在数?P中的位置
    T[t].l=makeKDTree(l,mid,depth+1);//生出左子?
    T[t].r=makeKDTree(mid+1,r,depth+1);//生出右子?

    return t;//返回?位置在P中的索引
}

void find(int v,int sx,int tx,int sy,int ty,int depth,vector<Point> &ans){
    int x=P[T[v].location].x;//v?点在数?P中的横坐?x
    int y=P[T[v].location].y;//v?点在数?P中的?坐?

    if(sx<=x&&x<=tx&&sy<=y&&y<=ty){
        ans.push_back(P[T[v].location]);//如果在范?内就放入答案数?中
    }

    if(depth%2==0){//此??按照x升序排序
        if(T[v].l!=NIL){
            if(sx<=x) find(T[v].l,sx,tx,sy,ty,depth+1,ans);
        }
        if(T[v].r!=NIL){
            if(x<=tx) find(T[v].r,sx,tx,sy,ty,depth+1,ans);
        }
    }
    else{
        if(T[v].l!=NIL)
            if(sy<=y) find(T[v].l,sx,tx,sy,ty,depth+1,ans);
        if(T[v].r!=NIL)
            if(y<=ty) find(T[v].r,sx,tx,sy,ty,depth+1,ans);
    }

}

int main(void){
    scanf("%d",&N);
    for(int i=0;i<N;i++){
        scanf("%d %d",&P[i].x,&P[i].y);
        P[i].id=i;
    }
    int q;
    root=makeKDTree(0,N,0);
    scanf("%d",&q);
    for(int j=0;j<q;j++){
        int sx,tx,sy,ty;
        scanf("%d %d %d %d",&sx,&tx,&sy,&ty);
        init();
        find(root,sx,tx,sy,ty,0,ans);
        sort(ans.begin(),ans.begin()+ans.size());
        for(int i=0;i<ans.size();i++)
            printf("%d\n",ans[i].id);
        printf("\n");
    }
    return 0;
}