We have N irregular jigsaw pieces. Each piece is composed of three rectangular parts of width 1 and various heights joined together. More specifically:

* The i-th piece is a part of height H, with another part of height A_i joined to the left, and yet another part of height B_i joined to the right, as shown below. Here, the bottom sides of the left and right parts are respectively at C_i and D_i units length above the bottom side of the center part.



<image>

Snuke is arranging these pieces on a square table of side 10^{100}. Here, the following conditions must be held:

* All pieces must be put on the table.
* The entire bottom side of the center part of each piece must touch the front side of the table.
* The entire bottom side of the non-center parts of each piece must either touch the front side of the table, or touch the top side of a part of some other piece.
* The pieces must not be rotated or flipped.



Determine whether such an arrangement is possible.

Constraints

* 1 \leq N \leq 100000
* 1 \leq H \leq 200
* 1 \leq A_i \leq H
* 1 \leq B_i \leq H
* 0 \leq C_i \leq H - A_i
* 0 \leq D_i \leq H - B_i
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N H
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_N B_N C_N D_N


Output

If it is possible to arrange the pieces under the conditions, print `YES`; if it is impossible, print `NO`.

Examples

Input

3 4
1 1 0 0
2 2 0 1
3 3 1 0


Output

YES


Input

4 2
1 1 0 1
1 1 0 1
1 1 0 1
1 1 0 1


Output

NO


Input

10 4
1 1 0 3
2 3 2 0
1 2 3 0
2 1 0 0
3 2 0 2
1 1 3 0
3 2 0 0
1 3 2 0
1 1 1 3
2 3 0 0


Output

YES
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=500;
int f[N],t[N],p[N],du[N];
int get(int x)
{
	if (f[x]==x) return x;
	return f[x]=get(f[x]);
}
int main()
{
	int n,h;
	scanf("%d%d",&n,&h);
	for (int i=1;i<=400;i++) f[i]=i;
	for (int i=1;i<=n;i++)
	{
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		int l=200,r=200;
		if (c) l=l-c;else l=l+a;
		if (d) r=r+d;else r=r-b;
		du[l]++;
		du[r]--;
		t[l]=t[r]=1;
		f[get(l)]=get(r);
	}
	for (int i=1;i<=200;i++) if (du[i]>0) 
	{
		puts("NO");
		return 0;
	}
	for (int i=201;i<=400;i++) if (du[i]<0)
	{
		puts("NO");
		return 0;
	}
	for (int i=1;i<=400;i++) 
	{
		if (du[i]) p[get(i)]=1;
		t[get(i)]=t[get(i)]|t[i];
	}
	for (int i=0;i<=400;i++) if (i==get(i)&&t[i]&&!p[i]) 
	{
		puts("NO");
		return 0;
	}
	 puts("YES");
	 return 0;
}