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 ll long long
#define cri const register int
#define re register
#define END return puts("NO"),0;
using namespace std;
int in[410],out[410],f[410],n,H,c[410],v[410];
int getf(cri x){ return f[x]==x?x:f[x]=getf(f[x]); }
int main(){
	cin>>n>>H;
	for(int i=1;i<=H+H;i++) f[i]=i;
	for(int i=1,a,b,c,d,L,R;i<=n;i++){
		cin>>a>>b>>c>>d;
		if(c) L=-c;else L=a;
		if(d) R=d;else R=-b;
		L+=H;R+=H;v[L]=v[R]=1;
		in[R]++;out[L]++;
		f[getf(L)]=getf(R);
	}
	for(int i=1;i<=H;i++) if(in[i]<out[i]) END;
	for(int i=H+1;i<=H+H;i++) if(in[i]>out[i]) END;
	for(int i=1;i<=H+H;i++) c[getf(i)]|=in[i]!=out[i];
	for(int i=1;i<=H+H;i++) if(v[i]&&!c[getf(i)]) END;
	puts("YES");
}