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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=405;

int n,h,a,b,c,d,in[N],out[N],f[N];
bool ok[N],vis[N];

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

void merge(int x,int y) {
	x=get(x);y=get(y);
	if (x==y) return;
	f[y]=x;
}

int main() {
	scanf("%d%d",&n,&h);
	fo(i,1,n) {
		scanf("%d%d%d%d",&a,&b,&c,&d);
		int l=c?-c:a;l+=h;
		int r=d?d:-b;r+=h;
		in[r]++;out[l]++;
		vis[l]=vis[r]=1;
		merge(l,r);
	}
	fo(i,h+1,h*2) if (in[i]>out[i]) {puts("NO");return 0;}
	fo(i,0,h-1) if (in[i]<out[i]) {puts("NO");return 0;}
	fo(i,0,h*2) ok[get(i)]|=(in[i]!=out[i]);
	fo(i,0,h*2) if (vis[i]&&get(i)==i&&!ok[i]) {puts("NO");return 0;}
	puts("YES");
	return 0;
}