#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n,h,fa[440],pd[440],du[2][440];
int get(int x)
{
	return fa[x]==x?x:fa[x]=get(fa[x]);
}
int main()
{
	scanf("%d%d",&n,&h);
	for(int i=1;i<=h+h;i++) fa[i]=i;
	for(int i=1,a,b,c,d;i<=n;i++)
	{
		scanf("%d%d%d%d",&a,&b,&c,&d);
		if(c==0) du[0][h+a]++; else du[0][h-c]++;
		if(d==0) du[1][h-b]++; else du[1][h+d]++;
		fa[get(c==0?h+a:h-c)]=get(d==0?h-b:h+d);
	}
	for(int i=0;i<h;i++) 
		if(du[0][i]>du[1][i]) {puts("NO");return 0;} 
	for(int i=h+1;i<=h+h;i++)
		if(du[0][i]<du[1][i]) {puts("NO");return 0;}
	for(int i=0;i<=h+h;i++)
		if(du[0][i]!=du[1][i]||du[0][i]+du[1][i]==0) pd[get(i)]=1;
	for(int i=0;i<=h+h;i++)
		if(!pd[get(i)]) {puts("NO");return 0;}
	puts("YES"); return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

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

### Response

```cpp
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n,h,fa[440],pd[440],du[2][440];
int get(int x)
{
	return fa[x]==x?x:fa[x]=get(fa[x]);
}
int main()
{
	scanf("%d%d",&n,&h);
	for(int i=1;i<=h+h;i++) fa[i]=i;
	for(int i=1,a,b,c,d;i<=n;i++)
	{
		scanf("%d%d%d%d",&a,&b,&c,&d);
		if(c==0) du[0][h+a]++; else du[0][h-c]++;
		if(d==0) du[1][h-b]++; else du[1][h+d]++;
		fa[get(c==0?h+a:h-c)]=get(d==0?h-b:h+d);
	}
	for(int i=0;i<h;i++) 
		if(du[0][i]>du[1][i]) {puts("NO");return 0;} 
	for(int i=h+1;i<=h+h;i++)
		if(du[0][i]<du[1][i]) {puts("NO");return 0;}
	for(int i=0;i<=h+h;i++)
		if(du[0][i]!=du[1][i]||du[0][i]+du[1][i]==0) pd[get(i)]=1;
	for(int i=0;i<=h+h;i++)
		if(!pd[get(i)]) {puts("NO");return 0;}
	puts("YES"); return 0;
}
```