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>
#define ll long long
#define N 100005
#define For(i,x,y) for(int i=x;i<=y;i++)
#define Rep(i,x,y) for(int i=x;i>=y;i--)
#define mp make_pair
#define pa pair<ll,int>
using namespace std;
inline int read(){
    int x=0,f=0;char ch=getchar();
    for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=1;
    for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-48;
    return f?-x:x;
}
int n,h,in[405],ou[405],fa[405];
bool flag[405];
inline int find(int x){return (fa[x]==x)?x:fa[x]=find(fa[x]);}
int main(){
    n=read();h=read();
    For(i,1,h*2) fa[i]=i;
    For(i,1,n){
        int a=read(),b=read(),c=read(),d=read();
        int l=(c==0)?h+a:c;
        int r=(d==0)?b:h+d;
        in[r]++;ou[l]++;
        fa[find(l)]=find(r);
    }
    For(i,1,h) if(ou[i]>in[i]){puts("NO");return 0;}
    For(i,h+1,h*2) if(ou[i]<in[i]){puts("NO");return 0;}
    For(i,1,h*2) if(in[i]!=ou[i]) flag[find(i)]=1;
    For(i,1,h*2) if(find(i)==i&&!flag[i]&&in[i]+ou[i]!=0){puts("NO");return 0;}
    puts("YES");
    return 0;
}