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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;

int get(){
	char ch;
	while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
	if (ch=='-'){
		int s=0;
		while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
		return -s;
	}
	int s=ch-'0';
	while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
	return s;
}

const int H = 210;

int n,h;
int deg[H*2];
int cnt[H*2];
int fa[H*2],siz[H*2];
bool bz[H*2],sef[H*2];

int getfather(int x){
	if (fa[x]==x)return fa[x];
	return fa[x]=getfather(fa[x]);
}

int main(){
	n=get();h=get();
	fo(i,0,h*2)fa[i]=i,siz[i]=1;
	fo(i,1,n){
		int a=get(),b=get(),c=get(),d=get();
		if (c)a=-c;
		if (d)b=d;else b=-b;
		a+=h;b+=h;
		int ta=getfather(a),tb=getfather(b);
		deg[a]++;deg[b]--;
		if (ta!=tb){
			fa[ta]=tb;
			siz[tb]+=siz[ta];
		}
		else sef[ta]=1;
	}
	bool ans=1;
	fo(i,0,h)if (deg[i]>0)ans=0;
	fo(i,h,2*h)if (deg[i]<0)ans=0;
	fo(i,0,h*2)bz[getfather(i)]|=(deg[i]!=0);
	fo(i,0,h*2)
	if (siz[i]>1&&getfather(i)==i&&!bz[i])ans=0;
	fo(i,0,h*2)
	if (siz[i]==1&&sef[i]&&getfather(i)==i)ans=0;
	if (ans)printf("YES\n");
	else printf("NO\n");
	return 0;
}