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

using namespace std;
typedef long long ll;
const ll N=500;

int n,h;
int fa[N],ok[N],ou[N],into[N];

inline int read()
{
	int o=0; char ch=' ';
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar())o=o*10+ch-48;
	return o;
}

inline int gf(int o)
{return fa[o]==o?o:fa[o]=gf(fa[o]);}

int main()
{
	cin>>n>>h;
	int a,b,c,d;
	fo(i,1,h<<1)fa[i]=i;
	fo(i,1,n){
		a=read(),b=read(),c=read(),d=read();
		int zb=0,yb=0; 
		if(c)zb=c+h;else zb=a;
		if(d)yb=d;else yb=b+h;
		++ou[zb]; ++into[yb];
		fa[gf(yb)]=gf(zb);
	}
	fo(i,1,(h<<1))if(into[i]!=ou[i])ok[i]=1;
	fo(i,1,h)if(into[i]>ou[i]){
		puts("NO"); return 0;
	}
	fo(i,h+1,h<<1)if(into[i]<ou[i]){
		puts("NO"); return 0;
	}
	fo(i,1,h<<1)ok[gf(i)]|=ok[i];
	fo(i,1,h<<1)if((into[i]||ou[i])&&!ok[i]&&fa[i]==i){
		puts("NO"); return 0;
	}
	puts("YES");
}