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.



<image>

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<cstdio>
#include<vector>
using namespace std;
const int MAXN=100005,MAXH=405;

int N,H;
int in[MAXH],out[MAXH];
int cntS[MAXH],cntT[MAXH];
bool vis[MAXH];

int fa[MAXH];
int Root(int x)
{return fa[x]==0?x:(fa[x]=Root(fa[x]));}
void Union(int x,int y)
{
	int r1=Root(x),r2=Root(y);
	if(r1==r2)
		return;
	fa[r1]=r2;
}

int ID(int h,int d)
{return h*2-1+d;}

int main()
{
	scanf("%d%d",&N,&H);
	for(int i=1;i<=N;i++)
	{
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		int u,v;
		if(c==0)
			u=ID(a,0);
		else
			u=ID(c,1);
		if(d==0)
			v=ID(b,1);
		else
			v=ID(d,0);
		out[u]++;
		in[v]++;
		vis[u]=vis[v]=true;
		Union(u,v);
		//printf("(%d,%d)->(%d,%d)          %d->%d\n",(u-1)/2+1,(u&1)^1,(v-1)/2+1,(v&1)^1,u,v);
	}
	
	for(int i=1;i<=2*H;i++)
	{
		if(in[i]-out[i]>=1)
		{
			if(i&1)
			{
				puts("NO");
				return 0;
			}
			cntT[Root(i)]+=in[i]-out[i];
		}
		if(out[i]-in[i]>=1)
		{
			if((i&1)^1)
			{
				puts("NO");
				return 0;
			}
			cntS[Root(i)]+=out[i]-in[i];
		}
	}
	for(int i=1;i<=2*H;i++)
		if(vis[i]&&fa[i]==0&&(cntS[i]!=cntT[i]||(cntS[i]==0&&cntT[i]==0)))
		{
			puts("NO");
			return 0;
		}
	puts("YES");
	
	return 0;
}
