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<bits/stdc++.h>

using namespace std;

int n,m,fa[605],sz[605],in[605],out[605];
bool fl[605];
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void link(int u,int v){sz[v]++;if(u==v)return;fa[u]=v,sz[v]+=sz[u];}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<=2*m;i++)fa[i]=i;
	for(int i=1,a,b,c,d,l,r;i<=n;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		l=c?-c:a,r=d?d:-b;
		in[r+m]++,out[l+m]++;
		link(find(l+m),find(r+m));
	}
	for(int i=0;i<=2*m;i++){
		if(i<m&&in[i]<out[i])puts("NO"),exit(0);
		if(i>m&&in[i]>out[i])puts("NO"),exit(0);
		if(in[i]!=out[i])fl[find(i)]=true;
	}
	for(int i=0;i<=2*m;i++)if(sz[i]&&!fl[find(i)])puts("NO"),exit(0);
	puts("YES");
	return 0;
}