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;

#define N 120000

int n,m,in[N],out[N];
vector<int> g[N];
bool vis[N],fl;

void add(int x,int y){
	++out[x+=m]; ++in[y+=m];
	g[x].push_back(y); g[y].push_back(x);
}

void dfs(int u){
	vis[u]=1;
	fl|=in[u]!=out[u];
	if (u<=m&&in[u]<out[u]){puts("NO"); exit(0);}
	if (u>m&&in[u]>out[u]){puts("NO"); exit(0);}
	for (int v:g[u]) if (!vis[v]) dfs(v);
}

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i){
		int a,b,c,d; scanf("%d%d%d%d",&a,&b,&c,&d);
		int x=c?-c:a,y=d?d:-b;
		add(x,y);
	}
	for (int i=1;i<=m*2;++i)
		if (in[i]+out[i]&&!vis[i]){
			fl=0;
			dfs(i);
			if (!fl){puts("NO"); return 0;}
		}
	puts("YES");
	
	return 0;
}
			