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 maxm=2e5+10;
const int base=210;

int n,H,in[500],out[500],cnt;
bool diff[500],vis[500];
int tote,FIR[500],TO[maxm],NEXT[maxm];

void addedge(int u,int v)
{
//	printf("%d %d\n",u,v);
	u+=base; v+=base;
	out[u]++; in[v]++;

	TO[++tote]=v;
	NEXT[tote]=FIR[u];
	FIR[u]=tote;

	TO[++tote]=u;
	NEXT[tote]=FIR[v];
	FIR[v]=tote;
}

void dfs(int u)
{
	if (vis[u]) return;
	vis[u]=1;
	cnt+=diff[u];
	for (int p=FIR[u];p;p=NEXT[p])
		dfs(TO[p]);
}

int main()
{
#ifdef h10
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
#endif
	int i,a,b,c,d;
	scanf("%d%d",&n,&H);
	for (i=1;i<=n;i++)
	{
		scanf("%d%d%d%d",&a,&b,&c,&d);
		addedge(c?-c:a,d?d:-b);
	}

	for (i=-200;i<=200;i++)
		if (out[base+i]!=in[base+i])
		{
			diff[base+i]=1;
//			printf("%d : %d\n",i,out[base+i]-in[base+i]);
//			printf("%d\n",(i<0)^(out[base+i]<in[base+i]));
			if ((i<0)^(out[base+i]<in[base+i]))
			{
				puts("NO");
				return 0;
			}
		}

	for (i=-200;i<=200;i++)
		if (!vis[base+i]&&FIR[base+i])
		{
			cnt=0;
			dfs(base+i);
			if (!cnt)
			{
				puts("NO");
				return 0;
			}
		}
	puts("YES");
}
