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 S=(1<<20)+5;
char buf[S],*H,*T;
inline char Get()
{
    if(H==T) T=(H=buf)+fread(buf,1,S,stdin);
    if(H==T) return -1;return *H++;
}
inline int read()
{
    int x=0;char c=Get();
    while(!isdigit(c)) c=Get();
    while(isdigit(c)) x=x*10+c-'0',c=Get();
    return x;
}
const int N=410;
int n,h,in[N],out[N];
int fa[N];
bool ok[N];

int find(const int &x){return fa[x]==x?x:fa[x]=find(fa[x]);}

int main()
{
    n=read();h=read();
    for(int i=1;i<=(h<<1);i++) fa[i]=i;
    for(int i=1;i<=n;i++)
    {
        int A=read(),B=read(),C=read(),D=read();
        int x=C?C+h:A,y=D?D:B+h;
        out[x]++;in[y]++;
        x=find(x);y=find(y);
        if(x!=y) fa[x]=y;
    }
    for(int i=1;i<=h;i++) if(in[i]>out[i]) return !puts("NO");
    for(int i=h+1;i<=(h<<1);i++) if(in[i]<out[i]) return !puts("NO");
    for(int i=1;i<=(h<<1);i++) if(in[i]!=out[i]||!in[i]&&!out[i]) ok[find(i)]=1;
    for(int i=1;i<=(h<<1);i++) if(fa[i]==i&&!ok[i]) return !puts("NO");
    return !puts("YES");
}