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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define M 500
using namespace std;
int read(){
    int nm=0,fh=1; int cw=getchar();
    for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
    for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
    return nm*fh;
}
int n,m,e1,e2,h1,h2,cnt,in[M],ot[M],f[M]; bool vs[M];
int fd(int x){return x==f[x]?x:f[x]=fd(f[x]);}
int main(){
    n=read(),m=read();
    for(int i=1;i<=m;i++) f[i]=i,f[i+m]=i+m,vs[i]=vs[i+m]=false;
    for(int i=1;i<=n;i++){
        h1=read(),h2=read(),e1=read(),e2=read();
        int t1=e1>0?e1:-h1,t2=e2>0?-e2:h2; t1+=m,t2+=m;
        if(fd(t1)!=fd(t2)) f[fd(t1)]=fd(t2); in[t2]++,ot[t1]++;
    }
    for(int i=-m;i<0;i++) if(in[i+m]>ot[i+m]){puts("NO");return 0;}
    for(int i=1;i<=m;i++) if(in[i+m]<ot[i+m]){puts("NO");return 0;}
    for(int i=0;i<=m+m;i++) if(in[i]!=ot[i]) vs[fd(i)]=true;
    for(int i=0;i<=m+m;i++) if(in[i]&&ot[i]&&!vs[fd(i)]){puts("NO");return 0;}
    puts("YES"); return 0;
}