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.



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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<iostream> 
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#define eps 1e-10
#define inf 1e9
#define For(i,j,k)	for(int i=j;i<=k;++i)
#define Dow(i,j,k)	for(int i=k;i>=j;--i)
#define ll long long
#define y1 orz
using namespace std;
inline ll read()
{
	ll t=0,f=1;char c=getchar();
	while(c<'0'||c>'9')	{if(c=='-')	f=-1;c=getchar();}
	while(c>='0'&&c<='9')	t=t*10+c-48,c=getchar();
	return t*f;
}
int n,h,F[1001],deg[1001],hav[1001],is_father[1001];
inline int get(int x){return x==F[x]?F[x]:F[x]=get(F[x]);}
int x,y,x1,y1,tx,ty;
int main()
{
	n=read();h=read();
	For(i,1,1000)	F[i]=i;
	For(i,1,n)	
	{
		x=read();y=read();x1=read();y1=read();
		if(x1==0)	tx=x;else tx=-x1;
		if(y1==0)	ty=-y;else ty=y1;
		tx+=500;ty+=500;
		deg[tx]++;deg[ty]--;hav[tx]=1;
//		cout<<tx<<' '<<ty<<endl;
		F[get(tx)]=get(ty);
	}
	For(i,501,1000)	if(deg[i]<0)	{puts("NO");return 0;}		
	For(i,1,500)	if(deg[i]>0)	{puts("NO");return 0;}
	For(i,1,1000)	
		hav[get(i)]|=hav[i];
	For(i,1,1000)
		if(deg[i])is_father[get(i)]=1;
	For(i,1,1000)	
		if(hav[i])
			if(F[i]==i&&!is_father[i])	return puts("NO"),0;
	puts("YES");
}