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 <stdio.h>
#include <string.h>
int F() {
	register int a, b, c;
	while(c=getchar(), c<'-');
	b=(c=='-');
	for(a=b?0:c-'0'; c=getchar(),c>'-'; a=a*10+c-'0');
	return b?-a:a;
}
const int N=2e5+10;
int fa[N];
int find(register int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
int deg[N];
bool mark[N], appear[N];
bool work() {
	register int n=F(), h=F(), i, a, b, c, d, cnt=0, ans=0;
	for(i=1; i<=2*h; ++i) fa[i]=i, deg[i]=0, appear[i]=mark[i]=0;
	for(i=1; i<=n; ++i) {
		a=F(), b=F(), c=F(), d=F();
		++deg[c=c?c:a+h];
		--deg[d=d?d+h:b];
		fa[find(c)]=find(d);
		appear[c]=appear[d]=1;
	}
	for(i=1; i<=2*h; ++i) if(appear[i]) {
		if(i<=h&&deg[i]>0 || i>h&&deg[i]<0) return 0;
		mark[i] = deg[i]!=0;
	}
	for(i=1; i<=2*h; ++i) mark[find(i)]|=mark[i];
	for(i=1; i<=2*h; ++i) if(appear[i] && find(i)==i && !mark[i]) return 0;
	return 1;
}
int main() {
	register int t=1;
	while(t--) puts(work()?"YES":"NO");
	return 0;
}