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
//Heaplax
#include<bits/stdc++.h>
#define LL long long
#define LOG(x) cerr<<#x<<" = "<<x<<endl
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;
}
using namespace std;
int n,h,in[405],out[405],f[405];
bool vis[405];
int find(int x)
{
	return f[x]==x?x:f[x]=find(f[x]);
}
int main()
{
	re(n),re(h);
	for(int i=1;i<=h*2;++i)f[i]=i;
	for(int i=1,a,b,c,d;i<=n;++i)
	{
		re(a),re(b),re(c),re(d);
		int l=0,r=0;
		if(c)l=c;else l=h+a;
		if(d)r=h+d;else r=b;
		if(find(l)!=find(r))f[f[l]]=f[r];
		++out[l];++in[r];
	}
	for(int i=1;i<=h;++i)if(out[i]>in[i])return puts("NO"),0;
	for(int i=h+1;i<=h+h;++i)if(out[i]<in[i])return puts("NO"),0;
	for(int i=1;i<=h+h;++i)if(in[i]!=out[i])vis[find(i)]=1;
	for(int i=1;i<=h+h;++i)if(in[i]&&out[i]&&!vis[find(i)])return puts("NO"),0;
	puts("YES");
}
