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 <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <ctime>
#include <map>
#include <queue>
#include <cstdlib>
#include <string>
#include <climits>
#include <set>
#include <vector>
using namespace std;
inline int read(){
	int k=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){k=k*10+ch-'0';ch=getchar();}
	return k*f;
}
int n,m,fa[200010],du[200010],vis[200010],c[200010];
inline int getfather(int x){return fa[x]==x?x:fa[x]=getfather(fa[x]);}
int main()
{
	n=read();m=read();
	for(int i=1;i<=1000;i++)fa[i]=i;
	for(int i=1;i<=n;i++){
		int a=read(),b=read(),c=read(),d=read();
		int x=(c==0)?a:-c,y=(d==0)?-b:d;
		x+=500;y+=500;int fx=getfather(x),fy=getfather(y);
		fa[fx]=fy;vis[x]=1;du[x]++;du[y]--;
	}
	bool flag=1;
	for(int i=1;i<500;i++)if(du[i]>0)flag=0;
	for(int i=501;i<=1000;i++)if(du[i]<0)flag=0;
	for(int i=1;i<=1000;i++){
		int fi=getfather(i);
		vis[fi]|=vis[i];
		if(du[i])c[fi]=1;
	}
	for(int i=1;i<=1000;i++)if(fa[i]==i&&!c[i]&&vis[i])flag=0;
	puts(flag?"YES":"NO");
	return 0;
}