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<stdio.h>
int N,H,Q;
int ace[444],in[444],out[444];
int ancestor(int x){return x-ace[x]?ace[x]=ancestor(ace[x]):x;}
bool proc(){
	for(int i=1; i<=H; i++)
		if(out[i] < in[i])return false;
	for(int i=H+1; i<=2*H; i++)
		if(out[i] > in[i])return false;
	for(int i=1; i<=2*H; i++)if(ancestor(i) == i){
		bool val = false, valid = false;
		for(int j=1; j<=2*H; j++)if(ancestor(j) == i){
			if(in[j] == 0 && out[j] == 0)continue;
			val = true;
			if(in[j] != out[j])valid = true;
		}
		if(val && !valid)return false;
	}
	return true;
}
int main(){
	scanf("%d%d",&N,&H);
	for(int i=1; i<=2*H; i++)
		ace[i] = i;
	Q = 2 * H + 1;
	for(int i=0; i<N; i++){
		int A,B,C,D,L,R;
		scanf("%d%d%d%d",&A,&B,&C,&D);
		if(C == 0)L = A;else
			L = Q - C;
		if(D == 0)R = Q - B;else
			R = D;
		in[R] ++;
		out[L] ++;
		ace[ancestor(L)] = ancestor(R);
	}
	//(l1,r1)-(l2,r2) => r1 == l2 || r1 < 0 && l2 > 0
	puts(proc()?"YES":"NO");
	return 0;
}