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
//此程序使用了qt专属名称，请勿模仿。侵权必究！！！
//我真是太菜了啊……
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=410;
int n,m,fa[N],in[N],out[N];
bool v[N];
inline int read()
{
	int sum=0,flag=1; char c;
	for(;c<'0'||c>'9';c=getchar()) if(c=='-') flag=-1;
	for(;c>='0'&&c<='9';c=getchar()) sum=(sum<<3)+(sum<<1)+c-'0';
	return sum*flag;
}
inline int getfa(int x)
{
	if(fa[x]^x) fa[x]=getfa(fa[x]);
	return fa[x];
}
int main()
{
	n=read(),m=read();
	for(int i=1;i<=m*2;++i) fa[i]=i;
	for(int i=1;i<=n;++i)
	{
		int a=read(),b=read(),c=read(),d=read(),x,y;
		if(c) x=m-c; else x=m+a;
		if(d) y=m+d; else y=m-b;
		++in[y],++out[x];
		fa[getfa(x)]=getfa(y);
	}
	for(int i=1;i<=m;++i) if(in[i]<out[i]) {puts("NO"); return 0;}
	for(int i=m+1;i<=m*2;++i) if(in[i]>out[i]) {puts("NO"); return 0;}
	for(int i=1;i<=m*2;++i) v[getfa(i)]|=(in[i]^out[i])||(in[i]==0&&out[i]==0);
	for(int i=1;i<=m*2;++i) if((fa[i]==i)&&(!v[i])&&(i^m)) {puts("NO"); return 0;}
	puts("YES"); return 0; 
}