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>
#define rep(i,j,k) for (i=j;i<=k;i++)
using namespace std;
const int N=1e5+5,H=405;
int n,i,u,v,I[H],O[H],fa[H],ok[H],a[N],b[N],c[N],d[N],h;
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]); }
void link(int u,int v) {
	int fx=find(u),fy=find(v);
	if (fx==fy) return;
	fa[fy]=fx;
}
int main()
{
//	freopen("jigsaw.in","r",stdin);
//	freopen("jigsaw.out","w",stdout);
	scanf("%d%d",&n,&h);
	rep(i,1,n) scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
	rep(i,1,h+h) fa[i]=i;
	rep(i,1,n)
	{
		if (!c[i]) u=a[i];
		else u=c[i]+h;
		if (!d[i]) v=b[i]+h;
		else v=d[i];
		O[u]++; I[v]++;
		link(u,v);
	}
	rep(i,1,h)
		if (O[i]<I[i]) { printf("NO\n"); return 0; }
	rep(i,h+1,h+h)
		if (O[i]>I[i]) { printf("NO\n"); return 0; }
	rep(i,1,h+h) if (O[i]!=I[i]) ok[find(i)]=1;
	rep(i,1,h+h)
		if (I[i] && !ok[find(i)]) { printf("NO\n"); return 0; }
	printf("YES\n");
	return 0;
}