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>
using namespace std;

const int MAXN=405;

int n,h;
int in[MAXN],out[MAXN],cnt[MAXN],fa[MAXN];

int Find(int x){return fa[x]==0?x:fa[x]=Find(fa[x]);}

int main(){
	scanf("%d%d",&n,&h);
	for(int i=1;i<=n;i++){
        int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		int x=c==0?a:(c+200);
		int y=d==0?(b+200):d;
		in[y]++,out[x]++;
		x=Find(x),y=Find(y);
		if(x!=y) fa[x]=y;
	}
	bool f=1;
	for(int i=1;i<=400;i++)
		if(in[i]!=out[i]){
			if((in[i]<out[i]&&i>200)||(in[i]>out[i]&&i<200)){
				f=0;
				break;
			}
			int x=Find(i);
			cnt[x]++;
		}
	if(f){
		for(int i=1;i<=400;i++)
			if((in[i]||out[i])&&cnt[Find(i)]==0){
				f=0;
				break;
			}
	}
	if(f) puts("YES");
	else puts("NO");
}
