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>
using namespace std;
int in[402],fa[402];
bool vis[402],qwq[402];
int h;
int getf(int x){
	return fa[x]==x?x:fa[x]=getf(fa[x]);
}
void add(int l,int r){
	l+=h;
	r+=h;
	in[r]++;
	in[l]--;
	qwq[l]=qwq[r]=1;
	int x=getf(l),y=getf(r);
	fa[x]=y;
}
int main(){
	int n;
	scanf("%d%d",&n,&h);
	for(int i=0;i<=2*h;i++)
		fa[i]=i;
	while(n--){
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		add(c?-c:a,d?d:-b);
	}
	for(int i=0;i<h;i++)
		if (in[i]<0)
			return puts("NO"),0;
	for(int i=h+1;i<=2*h;i++)
		if (in[i]>0)
			return puts("NO"),0;
	//puts("WTF");
	for(int i=0;i<=2*h;i++)
		if (in[i])
			vis[getf(i)]=1;
	//puts("WTF");
	for(int i=0;i<=2*h;i++)
		if (qwq[i] && !vis[getf(i)])
			return puts("NO"),0;
	puts("YES");
}