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<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#include<queue>
#include<bitset>
#include<map>
#include<set>

#define maxn 200005
#define MOD 1000000007

using namespace std;

inline int getint()
{
	int num=0,flag=1;char c;
	while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
	while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
	return num*flag;
}

int n,H;
int f[maxn];
int In[maxn],Ot[maxn];
int vis[maxn];

inline int find(int x)
{return f[x]==x?x:f[x]=find(f[x]);}
int main()
{
	n=getint(),H=getint();
	for(int i=1;i<=2*H;i++)f[i]=i;
	for(int i=1;i<=n;i++)
	{
		int A=getint(),B=getint(),C=getint(),D=getint();
		int x=C?C+H:A,y=D?D:B+H;
		Ot[x]++,In[y]++;
		x=find(x),y=find(y);
		if(x!=y)f[y]=x;
	}
	for(int i=1;i<=H;i++)if(In[i]>Ot[i]){printf("NO\n");return 0;}
	for(int i=H+1;i<=2*H;i++)if(In[i]<Ot[i]){printf("NO\n");return 0;}
	for(int i=1;i<=2*H;i++)if((!In[i]&&!Ot[i])||In[i]!=Ot[i])vis[find(i)]=1;
	for(int i=1;i<=2*H;i++)if(find(i)==i&&!vis[i]){printf("NO\n");return 0;}
	printf("YES\n");
}