818C.soft thief

Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?

Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!

The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.

Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.

Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.

The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.

Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.

Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.

Input

The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.

The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.

Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1)  ≠  (x2, y2) and no cell is covered by more than one sofa.

The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).

Output

Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.

Examples

input

2

3 2

3 1 3 2

1 2 2 2

1 0 0 1

output

1

input

3

10 10

1 2 1 1

5 5 6 5

6 4 5 4

2 1 2 0

output

2

input

2

2 2

2 1 1 1

1 2 2 2

1 0 0 0

output

-1

Note

Let's consider the second example.

The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).

The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.

The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.

So the second one corresponds to the given conditions.

In the third example

The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.

The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.

And there is no sofa with the set (1, 0, 0, 0) so the answer is -1.

题意，沙发店里找一个符合上面，下面，左面，右面有几个沙发的沙发，注意，如果两个沙发坐标是(1,1)(2,1)和(1,4)(2,4),那么他们两个的左面和右面各有一个沙发

解法，四个方向分别排序，然后暴力跑每一个沙发，排序方向上坐标符合第几个沙发的就加一，然后最后找找有没有加到4的沙发就可以了，但是有一个要特判，就是如果，符合的太多了，在排序之后还有符合的，且他的排序方向上的两个坐标是不一样的，那么就不存在，比如(1,1)(2,1)和(1,4)(2,4)，要你找上面一个，下面0个左边一个，右边0 个的，明显，这时候的右边，都是1个，要输出-1

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#define sf scanf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define vi vector<int>
#define mp make_pair
#define pf printf
#define prf(x) printf("%d\n",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (ll i=a;i<n;i++)
#define per(i,a,n) for (ll i=a;i>=n;i--)
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const double eps=1e-6;
const double pi=acos(-1.0);
const int inf=0x7fffffff;
const int N=1e5+7;
struct node
{
	int id;
	int x1,y1,x2,y2;
}a[N];
void predeal(node &x)
{
	if(x.x1==x.x2)
	{
		if(x.y1>x.y2)
		swap(x.y1,x.y2);
	}
	if(x.y1==x.y2)
	{
		if(x.x1>x.x2)
		swap(x.x1,x.x2);
	}
}
bool cmp1(node x,node y)
{
	if(x.x2!=y.x2)
	return x.x2 <y.x2;
	return x.x1 <y.x1;
}
bool cmp2(node x,node y)
{
	if(x.x1 !=y.x1)
	return x.x1 >y.x1;
	return x.x2 >y.x2;
 }
bool cmp3(node x,node y)
{
	if(x.y2!=y.y2)
	return x.y2 <y.y2 ;
	return x.y1 <y.y1 ;
}
bool cmp4(node x,node y)
{
	if(x.y1 !=y.y1 )
	return x.y1 >y.y1;
	return x.y2 >y.y2;
}
int num[N];
int main()
{
	mm(num,0);
	int q;scf(q);
	int n,m;scff(n,m);
	rep(i,1,1+q)
	{
		a[i].id=i;
		scff(a[i].x1,a[i].y1);
		scff(a[i].x2,a[i].y2);
		predeal(a[i]);
	}
	int u,d,l,r;
	cin>>l>>r>>d>>u;
	sort(a+1,a+q+1,cmp1);
	rep(i,1,q+1)
	{
		if(a[l+1].x1==a[i].x1&&a[l+1].x2==a[i].x2)
		{
			num[a[i].id]++;
			if(i>l+1&&a[i].x1 !=a[i].x2)//在排序方向要不是同一个高度，这样才会加一，如果是同一个高度那么就符合
			{
				cout<<"-1";return 0;
			}
		}
	}
	sort(a+1,a+1+q,cmp2);
	rep(i,1,q+1)
	{
		if(a[r+1].x1==a[i].x1&&a[r+1].x2==a[i].x2)
		{
			num[a[i].id]++;
			if(i>r+1&&a[i].x2!=a[i].x1)
			{
				cout<<"-1";return 0;
			}
		}
	}
	sort(a+1,a+1+q,cmp3);
	rep(i ,1,q+1)
	{
		if(a[d+1].y1==a[i].y1&&a[d+1].y2==a[i].y2)
		{
			num[a[i].id]++;
			if(i>d+1&&a[i].y1!=a[i].y2)
			{
				cout<<"-1";return 0;
			}
		}
	}
	sort(a+1,a+1+q,cmp4);
	rep(i,1,q+1)
	{
		if(a[u+1].y1==a[i].y1&&a[u+1].y2==a[i].y2)
		{
			num[a[i].id]++;
			if(i>u+1&&a[i].y1 !=a[i].y2)
			{
				cout<<"-1";return 0;
			}
		}

	}
	int ans=-1;
	rep(i,1,q+1)
	if(num[i]==4) ans=i;
	prf(ans);
	return 0;
}