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<cstdio>
#include<algorithm>
#define NO return puts("NO"),0
using namespace std;
 
int n,m,f[1000],a,b,c,d,x,y,D[1000],bo[1000],C[1000];
int gf(int x){return x==f[x]?x:f[x]=gf(f[x]);}
int main(){
    scanf("%d%d",&n,&m);
    
    for (int i=1;i<999;i++) f[i]=i;
    while (n--){
        scanf("%d%d%d%d",&a,&b,&c,&d);
        if (c==0) x=a;else x=-c;x+=500;
        if (d==0) y=-b;else y=d;y+=500;
        f[gf(x)]=gf(y);
        D[x]++;D[y]--;bo[x]=1;
    }
    for (int i=1;i<500;i++) if (D[i]>0) NO;
    for (int i=501;i<999;i++) if (D[i]<0) NO;
    for (int i=1;i<999;i++)
    if (bo[gf(i)]|=bo[i],D[i]) C[f[i]]=1;
    for (int i=1;i<999;i++)
    if (f[i]==i&&!C[i]&&bo[i]) NO;
    puts("YES");
}