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

const int N=100100;
int fa[N],f[N],g[N],n,m,vis[N];
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=0;i<=2*m;++i)fa[i]=i;
	for(int i=1;i<=n;++i){
		int a,b,c,d;cin>>a>>b>>c>>d;
		int u=c?m+c:m-a,v=d?m-d:m+b;
		fa[find(u)]=find(v);f[u]++;g[v]++;
	}
	for(int i=1;i<=m;++i)
		if(f[m-i]<g[m-i]||f[m+i]>g[m+i]){cout<<"NO\n";return 0;}
	for(int i=0;i<=2*m;++i)if(f[i]!=g[i])vis[find(i)]=1;
	for(int i=0;i<=2*m;++i)if(f[i]&&g[i]&&!vis[find(i)]){cout<<"NO\n";return 0;}
	cout<<"YES\n";
	return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int N=100100;
int fa[N],f[N],g[N],n,m,vis[N];
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=0;i<=2*m;++i)fa[i]=i;
	for(int i=1;i<=n;++i){
		int a,b,c,d;cin>>a>>b>>c>>d;
		int u=c?m+c:m-a,v=d?m-d:m+b;
		fa[find(u)]=find(v);f[u]++;g[v]++;
	}
	for(int i=1;i<=m;++i)
		if(f[m-i]<g[m-i]||f[m+i]>g[m+i]){cout<<"NO\n";return 0;}
	for(int i=0;i<=2*m;++i)if(f[i]!=g[i])vis[find(i)]=1;
	for(int i=0;i<=2*m;++i)if(f[i]&&g[i]&&!vis[find(i)]){cout<<"NO\n";return 0;}
	cout<<"YES\n";
	return 0;
}

```