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<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=1e5+3,mh=203;
int n,h,gra,beg[mh*2],to[mn],nex[mn],vis[mh*2],ind[mh*2],outd[mh*2];
void ins(int u,int v){
	to[++gra]=v;
	nex[gra]=beg[u];
	beg[u]=gra;
}
void dfs(int p){
	vis[p]=1;
	fr(i,p)if(!vis[to[i]])dfs(to[i]);
}
int main(){
	scanf("%d%d",&n,&h);
	fo(i,1,n){
		int a,b,c,d,x,y;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		if(c)x=-c;else x=a;
		if(d)y=d;else y=-b;
		x+=h;y+=h;
		ins(x,y);
		ind[y]++;
		outd[x]++;
	}
	fo(i,0,h)if(outd[i]>ind[i]){printf("NO");return 0;}
	fo(i,h,h*2)if(outd[i]<ind[i]){printf("NO");return 0;}
	fo(i,0,h*2)if((ind[i]!=outd[i])&&(!vis[i]))dfs(i);
	fo(i,0,h*2)if((ind[i]||outd[i])&&(!vis[i])){printf("NO");return 0;}
	printf("YES");
	return 0;
}
