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<bits/stdc++.h>
#define N 100005
#define M 505
using namespace std;
int n,m,a[N],b[N],c[N],d[N],deg[M],f[M],vis[M],fr[M];
void GG(){ printf("NO"); exit(0);}
int find(int x){ return f[x] ? f[x]=find(f[x]) : x;}
int main()
{
	int i,x,y;
	scanf("%d %d",&n,&m),m=2*m+1;
	for(i=1;i<=n;i++){
		scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
		if(c[i]!=0) x=2*c[i]+1; else x=2*a[i];
		if(d[i]!=0) y=2*d[i]; else y=2*b[i]+1;
		deg[x]++,deg[y]--,vis[x]=vis[y]=1;
		if(find(x)!=find(y)) f[find(x)]=find(y);
	  }
	for(i=1;i<=m;i++)
		if(deg[i]!=0){
			if(deg[i]>0&&(i&1)) GG();
			if(deg[i]<0&&(~i&1)) GG();
		  }
	for(i=1;i<=m;i++)
		if(vis[i]&&deg[i]>0) fr[find(i)]++;
	for(i=1;i<=m;i++)
		if(vis[i]&&find(i)==i&&!fr[i]) GG();
	printf("YES");
	return 0;
}
