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<cstdio>
#include<cstring>
#define nt(x) (x+h)
namespace Hao_Nan_A___Wo_Bu_Hui_A___Wo_Hao_Ruo_A
{
	const int N=101000,M=510;
	int ind[M],outd[M],q[N];
	int n,h,e;
	int ask(int p){return p==q[p]?p:q[p]=ask(q[p]);}
	void link(int u,int v){q[ask(u)]=ask(v);}
	void add(int x,int y)
	{
//		printf("%d -> %d\n",x,y);
		ind[y]++;outd[x]++;
		link(x,y);
	}
	void initialize()
	{
		scanf("%d%d",&n,&h);e=0;
		for(int i=1;i<=h*2;i++)
			ind[i]=outd[i]=0,q[i]=i;
		for(int i=1,A,B,C,D,u,v;i<=n;i++)
		{
			scanf("%d%d%d%d",&A,&B,&C,&D);
			if(C)u=nt(C);else u=A;
			if(D)v=D;else v=nt(B);
			add(u,v);
		}
	}
	bool check()
	{
		static bool vis[M];
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=h;i++)if(ind[i]>outd[i])return 0;
		for(int i=1;i<=h;i++)if(ind[nt(i)]<outd[nt(i)])return 0;

		for(int i=1;i<=h<<1;i++)
			if(ind[i]!=outd[i])vis[ask(i)]=1;
		for(int i=1;i<=h<<1;i++)
			if(!vis[ask(i)] && (ind[i]||outd[i]))return 0;
		return 1;
	}
	void solve()
	{
		initialize();
		if(check())printf("YES\n");
		else printf("NO\n");
	}
}
int main()
{
//	freopen("in","r",stdin);
	Hao_Nan_A___Wo_Bu_Hui_A___Wo_Hao_Ruo_A::solve();
	return 0;
}
