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
//神题
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 410
using namespace std;
int n, m, du[N], f[N], size[N], flag[N], A, B, C, D, x, y, x1, y1;
inline int gether(int x){return f[x]?f[x]=gether(f[x]):x;}
int main(){
	scanf("%d%d", &n, &m); m<<=1;
	memset(du, 0, sizeof(du)); memset(f, 0, sizeof(f)); memset(size, 0, sizeof(size));
	for(int i=1; i<=n; i++){
		scanf("%d%d%d%d", &A, &B, &C, &D);
		x=C?-C:A; y=D?D:-B; if(x<=0)x+=m+1; if(y<=0)y+=m+1;
		du[x]++; du[y]--;
		x1=gether(x); y1=gether(y); if(x1!=y1){f[y1]=x1; size[x1]+=size[y1];} size[x1]++;
	}
	for(int i=1; i<=m/2; i++)if(du[i]<=-1){printf("NO"); return 0;}
	for(int i=m/2+1; i<=m; i++)if(du[i]>=1){printf("NO"); return 0;}
	memset(flag, 0, sizeof(flag));
	for(int i=1; i<=m; i++)if(du[i])flag[gether(i)]=1;
	for(int i=1; i<=m; i++)if(!f[i]&&size[i]&&!flag[i]){printf("NO"); return 0;}
	printf("YES");
	return 0;
}