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 <bits/stdc++.h>
using namespace std;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1  : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=4e2+50, H=201;
int n,d[N],vis[N],ok[N];
vector <int> edge[N];
inline void add(int x,int y) {
	x+=H; y+=H; ok[x]=ok[y]=1;
	--d[x]; ++d[y]; 
	edge[x].push_back(y);
	edge[y].push_back(x);
}
int tag;
inline bool dfs(int x) {
	if(x<=H && d[x]>0) return false;
	if(x>H && d[x]<0) return false;
	vis[x]=1; tag|=(d[x]!=0);
	for(auto v:edge[x])
		if(!vis[v]) if(!dfs(v)) return false;
	return true;
}
int main() {
	n=rd(); rd();
	for(int i=1;i<=n;i++) {
		int a=rd(), b=rd(), c=rd(), d=rd();
		add(c?c:-a,d?-d:b);
	} 
	for(int i=1;i<=2*H;i++) if(ok[i] && !vis[i]) {
		tag=0;
		if((!dfs(i)) || (!tag)) return puts("NO"),0;
	} puts("YES");
}