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>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=200000+10;
int fa[maxn],vis[maxn],du[maxn],bz[maxn];
int i,j,k,l,t,n,m,a,b,c,d,x,y;
bool czy;
int getfa(int x){
	return fa[x]==x?x:fa[x]=getfa(fa[x]);
}
int main(){
	scanf("%d%d",&n,&m);
	fo(i,1,1000) fa[i]=i;
	fo(i,1,n){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		x=c?-c:a;
		y=d?d:-b;
		x+=500;y+=500;
		fa[getfa(x)]=getfa(y);
		du[x]++;du[y]--;
		vis[x]=1;
	}
	czy=1;
	fo(i,1,499)
		if (du[i]>0) czy=0;
	fo(i,501,1000)
		if (du[i]<0) czy=0;
	fo(i,1,1000) vis[getfa(i)]|=vis[i];
	fo(i,1,1000)
		if (du[i]) bz[getfa(i)]=1;
	fo(i,1,1000)
		if (fa[i]==i&&!bz[i]&&vis[i]) czy=0;
	if (czy) printf("YES\n");else printf("NO\n");
}