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<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<string>
#include<stack>
#include<queue>

using namespace std;

struct pNode
{
    int id,x,y;
    pNode(){}//必不可少
    pNode(int id_,int x_,int y_):id(id_),x(x_),y(y_){}
};
struct Node
{
    int depth;
    pNode data;
    Node *l,*r;
};
int n,sx,tx,sy,ty;
static const int maxn=500010;
int len;
int ans[maxn];
pNode g[maxn];


bool cmpx(pNode a,pNode b)
{
    return a.x<b.x;
}
bool cmpy(pNode a,pNode b)
{
    return a.y<b.y;
}
void build(Node *&p,int depth,int left,int right)
{
    if(left>right)
        return;
    int mid=(left+right)/2;
    if(depth%2==0)
        sort(g+left,g+right+1,cmpx);
    else
        sort(g+left,g+right+1,cmpy);
    p=(Node *)malloc(sizeof(Node));
    p->depth=depth;
    p->data=g[mid];
    p->l=NULL,p->r=NULL;
    build(p->l,depth+1,left,mid-1);
    build(p->r,depth+1,mid+1,right);
}

void find_range(Node *&p)
{
    if(p==NULL)
        return;
    int x=p->data.x;
    int y=p->data.y;
    if(x>=sx &&x<=tx &&y>=sy &&y<=ty)
    {
        ans[len++]=p->data.id;
    }
    if(p->depth%2==0)
    {
        if(x>=sx)
            find_range(p->l);
        if(x<=tx)
            find_range(p->r);
    }else
    {
        if(y>=sy)
            find_range(p->l);
        if(y<=ty)
            find_range(p->r);
    }
}
int main()
{
    int i,j,q;
    Node *tree=NULL;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d %d",&g[i].x,&g[i].y);
        g[i].id=i;
    }
    build(tree,0,0,n-1);
    scanf("%d",&q);
    for(i=0;i<q;i++)
    {
        scanf("%d%d%d%d",&sx,&tx,&sy,&ty);
        len=0;
        find_range(tree);
        sort(ans,ans+len);
        for(j=0;j<len;j++)
            printf("%d\n",ans[j]);
        printf("\n");
    }
    return 0;
}

