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<stdio.h>
#include<string.h>
using namespace std;
int n,h;
int l[220],r[220],fa[420],siz[420],L,R;
int find(int x){
	if(fa[x]==x)return x;
	return fa[x]=find(fa[x]);
}
int main(){
	scanf("%d%d",&n,&h);
	for(int i=1;i<=400;i++)fa[i]=i;
	for(int i=1;i<=n;i++){
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		if(c==0)l[a]++,L=a;
		else r[c]--,L=c+200;
		if(d==0)r[b]++,R=b+200;
		else l[d]--,R=d;
		if(find(L)==find(R)){
			siz[find(L)]++;
			continue;
		}
		siz[find(R)]+=siz[find(L)]+1;
		siz[find(L)]=0;
		fa[find(L)]=find(R);
	}
	int ans=0,cns=0;
	for(int i=1;i<=200;i++){
		if(l[i]<0||r[i]<0){
			printf("NO\n");
			return 0;
		}
		if(l[i])ans+=siz[find(i)],siz[find(i)]=0;
		if(r[i])ans+=siz[find(i+200)],siz[find(i+200)]=0;
	}
	if(ans!=n)printf("NO\n");
	else printf("YES\n");
}