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 <cstdlib>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define efo(i,q) for(int i=A[q];i;i=B[i][0])
using namespace std;
const int N=100500;
int read(int &n)
{
	char ch=' ';int q=0,w=1;
	for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());
	if(ch=='-')w=-1,ch=getchar();
	for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;
}
int m,n,ans;
int Bv[N][2];
int B[2*N][2],A[N],B0;
bool z[N];
void dfs(int q)
{
	z[q]=1;
	efo(i,q)if(!z[B[i][1]])dfs(B[i][1]);
}
int main()
{
	int q,w;
	read(n),read(m);
	fo(i,1,n)
	{
		int qa,qb,qc,qd;
		read(qa),read(qb),read(qc),read(qd);
		if(qc==0)
		{
			q=m+qa;
		}else q=qc;
		if(qd==0)w=qb;
		else w=m+qd;
		++Bv[q][0];++Bv[w][1];
		B[++B0][0]=A[q],A[q]=B0,B[B0][1]=w;
	}
	ans=0;q=0;
	fo(i,1,m)
	{
		if(Bv[i][0]>Bv[i][1])++ans;
		if(Bv[i][0]<Bv[i][1])++q;
	}
	if(ans||!q)return printf("NO\n"),0;
	q=0;
	fo(i,m+1,m*2)
	{
		if(Bv[i][0]<Bv[i][1])++ans;
		if(Bv[i][0]>Bv[i][1])++q;
	}
	if(ans||!q)return printf("NO\n"),0;
	fo(i,m+1,2*m)if(Bv[i][0]>Bv[i][1]&&!z[i])dfs(i);
	fo(i,1,m*2)if(A[i]&&!z[i])return printf("NO\n"),0 ;
	printf("YES\n");
	return 0;
}