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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
const double EPS=1e-6;
const double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
int IN[1100];
int OUT[1100];
int UF[1100];
int FIND(int a){
	if(UF[a]<0)return a;
	return UF[a]=FIND(UF[a]);
}
void UNION(int a,int b){
	a=FIND(a);b=FIND(b);if(a==b)return;UF[a]+=UF[b];UF[b]=a;
}
int oo[1100];
int ii[1100];
int main(){
	int a,b;scanf("%d%d",&a,&b);
	for(int i=0;i<b*2+1;i++)UF[i]=-1;
	for(int i=0;i<a;i++){
		int p,q,r,s;scanf("%d%d%d%d",&p,&q,&r,&s);
		int fr=-r;
		int to=-s;
		if(r==0)fr=p;
		if(s==0)to=q;
		to*=-1;
		IN[fr+b]++;
		OUT[to+b]++;
		UNION(fr+b,to+b);
	}
	for(int i=0;i<2*b;i++){
		if(IN[i]>OUT[i]&&i-b<0){printf("NO\n");return 0;}
		if(IN[i]<OUT[i]&&i-b>0){printf("NO\n");return 0;}
	}
	for(int i=1;i<=b;i++){
		if(IN[i+b]>OUT[i+b]){
			oo[FIND(i+b)]++;
		}
		if(IN[-i+b]<OUT[-i+b]){
			ii[FIND(-i+b)]++;
		}
	}
	for(int i=1;i<=b;i++){
		if(IN[i+b]+OUT[i+b]&&(oo[FIND(i+b)]==0||ii[FIND(i+b)]==0)){printf("NO\n");return 0;}
		if(IN[-i+b]+OUT[-i+b]&&(oo[FIND(-i+b)]==0||ii[FIND(-i+b)]==0)){printf("NO\n");return 0;}
		
	}
	printf("YES\n");
}