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<iostream>
#include<algorithm>
#include<cmath>
#include<stdio.h>
using namespace std;
int n,m;
int fa[2000003],vis[2000003],D[2000003],C[2000003];
inline int find(int x){
	if(fa[x]==x){
		return fa[x];
	}else{
		return fa[x]=find(fa[x]);
	}
}
int main(){
	cin>>n>>m;
    for(int i=1;i<=1000;i++){
    	fa[i]=i;
	}
    for(int i=1;i<=n;i++){
        int x,y,a,b,c,d;
        cin>>a>>b>>c>>d;
        x=!c?a+500:-c+500;
        y=!d?-b+500:d+500;
        fa[find(x)]=find(y);
        D[x]++,D[y]--;
		vis[x]=1;
    }
    int flag=1;
    for(int i=1;i<500;i++){
    	if(D[i]>0){
			puts("NO");
    		return 0;
		}
	}
    for(int i=501;i<=1000;i++){
    	if(D[i]<0){
    		puts("NO");
    		return 0;
		}
	}
    for(int i=1;i<=1000;i++){
    	vis[find(i)]|=vis[i];
	}
    for(int i=1;i<=1000;i++){
		if(D[i]!=0){
			C[find(i)]=1;
		}
	}
    for(int i=1;i<=1000;i++){
    	if(fa[i]==i&&!C[i]&&vis[i]){
    		puts("NO");
    		return 0;
		}
	}
	cout<<"YES"<<endl;
    return 0;
}