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;
int rt(int *uf,int i)
{
  return uf[i]<0?i:uf[i]=rt(uf,uf[i]);
}

void con(int *uf,int a,int b)
{
  a=rt(uf,a);
  b=rt(uf,b);
  if(a!=b){
    if(-uf[a]<-uf[b]){
      uf[a]=b;
    }
    else if(-uf[a]>-uf[b]){
      uf[b]=a;
    }
    else{
      uf[a]=b;
      uf[b]--;
    }
  }
}

int main()
{
  int n,h;
  scanf("%d%d",&n,&h);
  vector<int> A(n),B(n),C(n),D(n);
  for(int i=0;i<n;i++){
    scanf("%d%d%d%d",&(A[i]),&(B[i]),&(C[i]),&(D[i]));
  }
  vector<int> D0(h+1),D1(h+1);
  int uf[500],b[500]={0};
  for(int i=0;i<2*(h+1);i++){
    uf[i]=-1;
  }
  for(int i=0;i<n;i++){
    int x,y;
    if(C[i]==0){
      D0[A[i]]++;
      x=A[i];
    }
    else{
      D1[C[i]]--;
      x=C[i]+h+1;
    }
    if(D[i]==0){
      D1[B[i]]++;
      y=B[i]+h+1;
    }
    else{
      D0[D[i]]--;
      y=D[i];
    }
    con(uf,x,y);
    b[x]=b[y]=1;
  }
  bool F=0;
  for(int i=0;i<=h;i++){
    if(D0[i]<0||D1[i]<0){
      F=1;
    }
  }
  vector<int> B0(2*(h+1),0),B1(2*(h+1),0);
  for(int x=0;x<2*(h+1);x++){
    if(b[x]){
      //printf("%d %d\n",x,rt(uf,x));
      if(x<=h&&D0[x]>0){
	int r=rt(uf,x);
	B0[r]=1;
      }
      if(h<x&&D1[x-h-1]>0){
	int r=rt(uf,x);
	B1[r]=1;
      }
    }
  }
  for(int x=0;x<2*(h+1);x++){
    if(b[x]&&uf[x]<0&&(!B0[x]||!B1[x])){
      F=1;
    }
  }
  puts(F?"NO":"YES");
  return 0;
}
