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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int N=405;

typedef long long LL;

int n,h,de[N],f[N];

bool vis[N],bz[N];

int Get(int x)
{
	return (f[x]==x)?x:f[x]=Get(f[x]);
}

int main()
{
	for (int i=0;i<N;i++) f[i]=i;
	scanf("%d%d",&n,&h);
	for (int i=0,a,b,c,d,l,r;i<n;i++)
	{
		scanf("%d%d%d%d",&a,&b,&c,&d);
		l=(c==0)?a:-c; r=(d==0)?-b:d;
		l+=h; r+=h; vis[l]=vis[r]=1;
		de[l]++; de[r]--;
		a=Get(l); b=Get(r);
		f[b]=a;
	}
	for (int i=0;i<h;i++) if (de[i]>0)
	{
		printf("NO\n"); return 0;
	}
	for (int i=N-1;i>h;i--) if (de[i]<0)
	{
		printf("NO\n"); return 0;
	}
	for (int i=0;i<N;i++)
	{
		int j=Get(i); vis[j]|=vis[i]; bz[j]|=(de[i]!=0);
	}
	for (int i=0;i<N;i++) if (Get(i)==i && vis[i] && !bz[i])
	{
		printf("NO\n"); return 0;
	}
	printf("YES\n");
	return 0;
}