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>
int n,H,in[411],out[411];
bool e[411][411],d[411];
void dfs(int w){
	d[w]=1;
	for(int i=0;i<=H*2;++i)if(e[w][i]&&!d[i])dfs(i);
}
void ae(int x,int y){
	++out[x],++in[y];
	e[x][y]=1;
}
int main(){
	scanf("%d%d",&n,&H);
	for(int i=0,a,b,c,d;i<n;++i){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		ae(c?H+c:a,d?d:H+b);
	}
	for(int i=1;i<=H;++i){
		int w=i;
		if(in[w]>out[w])return puts("NO"),0;
		while(in[w]<out[w])ae(0,w);
		w=H+i;
		if(in[w]<out[w])return puts("NO"),0;
		while(in[w]>out[w])ae(w,0);
	}
	if(in[0]!=out[0])return puts("NO"),0;
	dfs(0);
	for(int i=1;i<=H*2;++i)if((in[i]|out[i])&&!d[i])return puts("NO"),0;
	return puts("YES"),0;
}