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>
const int MaxN=100010,MaxH=210;
int N,H;
struct edge{int to;edge*next;}E[MaxN*2],*ne=E,*fir[MaxH*2];
void link(int u,int v){*ne=(edge){v,fir[u]};fir[u]=ne++;}
int deg[MaxH*2],Q[MaxH*2];
bool vis[MaxH*2];
int main(){
	scanf("%d%d",&N,&H);
	for(int i=0;i<N;i++){
		int A,B,C,D;
		scanf("%d%d%d%d",&A,&B,&C,&D);
		A=C?C+H:A;
		B=D?D:B+H;
		link(A,B);link(B,A);
		deg[A]++;deg[B]--;
	}
	for(int i=1;i<=H*2;i++)if(!vis[i]&&fir[i]){
		bool fl=1,ex=0;
		int*h=Q,*t=Q;
		vis[*t++=i]=1;
		for(;h<t;h++){
			if(*h<=H&&deg[*h])ex=1,deg[*h]<0?fl=0:1;
			if(*h>H&&deg[*h])ex=1,deg[*h]>0?fl=0:1;
			for(edge*e=fir[*h];e;e=e->next)
				if(!vis[e->to])vis[*t++=e->to]=1;
		}
		if(!fl||!ex)return puts("NO"),0;
	}
	puts("YES");
}