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<iostream>
#include<cstdio>
using namespace std;
#define MAX 450
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,H,f[MAX],In[MAX],Ot[MAX];bool vis[MAX];
int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
int main()
{
	n=read();H=read();
	for(int i=1;i<=H+H;++i)f[i]=i,vis[i]=false;
	for(int i=1;i<=n;++i)
	{
		int a=read(),b=read(),c=read(),d=read();
		int l=(c>0?c:-a)+H,r=(d>0?-d:b)+H;
		if(getf(l)!=getf(r))f[getf(l)]=getf(r);
		In[r]+=1;Ot[l]+=1;
	}
	for(int i=-H;i<0;++i)if(In[i+H]>Ot[i+H]){puts("NO");return 0;}
	for(int i=1;i<=H;++i)if(In[i+H]<Ot[i+H]){puts("NO");return 0;}
	for(int i=0;i<=H+H;++i)if(In[i]!=Ot[i])vis[getf(i)]=true;
	for(int i=0;i<=H+H;++i)if(In[i]&&Ot[i]&&!vis[getf(i)]){puts("NO");return 0;}
	puts("YES");
	return 0;
}
