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.



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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>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int n,m,i,j,fa[200005],vis[200005];
int rd[200005],cd[200005];
int fnd(int x)
{
	if(fa[x]==x) return x;
	return fa[x]=fnd(fa[x]); 
}
int main()
{
	cin>>n>>m;m++;
	rep(i,m+m)fa[i]=i;
	rep(i,n)
	{
		int a,b,c,d,l,r;
		cin>>a>>b>>c>>d;
		if(c) l=m+c; else l=m-a;
		if(d) r=m-d; else r=m+b;
		rd[r]++;cd[l]++;
		fa[fnd(l)]=fnd(r);
	}
	rep(i,m) if(rd[i]>cd[i]) {puts("NO");return 0;}
	rep(i,m) if(rd[i+m]<cd[i+m]) {puts("NO");return 0;}
	rep(i,m+m) if(rd[i]!=cd[i]) vis[fnd(i)]=1;
	rep(i,m+m) if(cd[i]&&!vis[fnd(i)]) {puts("NO");return 0;}
	puts("YES");
	return 0;
}