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 beg[200100],nex[501000],tto[500100],e;
int cnt1[200100],cnt2[200100];
void putin(int s,int t){
	tto[++e]=t;
	nex[e]=beg[s];
	beg[s]=e;
}
bool vis[200100];
void dfs(int u){
	if(vis[u]) return;
	vis[u]=1;
	for(int i=beg[u];i;i=nex[i])
		dfs(tto[i]);
}
int main(){
//	freopen("E.in","r",stdin);
	int n,h;
	scanf("%d%d",&n,&h);
	int A,B,C,D;
	for(int i=1;i<=n;i++){
		scanf("%d%d%d%d",&A,&B,&C,&D);
		if(C>0){
			putin(n+C,i);
			putin(i,n+C);
			cnt1[n+C]++;
		}
		else{
			putin(n+h+A,i);
			putin(i,n+h+A);
			cnt2[n+A+h]++;
		}
		if(D>0){
			putin(n+D+h,i);
			putin(i,n+D+h);
			cnt1[n+D+h]++;
		}
		else{
			putin(n+B,i);
			putin(i,n+B);
			cnt2[n+B]++;
		}
	}
	for(int i=1;i<=h+h;i++){
		if(cnt1[n+i]>cnt2[n+i]){
			printf("NO\n");
			return 0;
		}
	}
	for(int i=1;i<=h+h;i++)
		if(cnt2[n+i]>cnt1[n+i])
			dfs(n+i);
	for(int i=1;i<=n;i++){
		if(!vis[i]){
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
}
