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<bits/stdc++.h>
using namespace std;
const int N=405;
int a,b,c,d,sd[N],n,h,Map[N][N],vis[N],ans=1,Flag;
void dfs(int x){
	if (!vis[x]||!ans)return;
	vis[x]=0;
	if (x>h&&sd[x]<0)ans=0;
	if (x<=h&&sd[x]>0)ans=0;
	if (sd[x])Flag=1;
	for (int i=1;i<=h*2;i++)
		if (Map[x][i])dfs(i);
}
int main(){
	scanf("%d%d",&n,&h);
	for (int i=1;i<=n;i++){
		scanf("%d%d%d%d",&a,&b,&c,&d);
		a=c?-c+h+1:a+h;b=d?d+h:-b+h+1;
		sd[a]++;sd[b]--;
		Map[a][b]=Map[b][a]=vis[a]=vis[b]=1;
	}
	for (int i=1;i<=h*2&&ans;i++)
		if (vis[i]){
			Flag=0;dfs(i);
			if (!Flag)ans=0;
		}
	if (ans)puts("YES");
	else puts("NO");
}
