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 maxn = 444;
int uf[maxn], has[maxn];
int in[maxn], out[maxn];
int find(int x){
	if(x == uf[x]) return x;
	return uf[x] = find(uf[x]);
}
int main(){
	iota(uf, uf + maxn, 0);
	int n, h; cin >> n >> h;
	for(int i = 0; i < n; i++){
		int a, b, c, d; cin >> a >> b >> c >> d;
		int x = (c == 0) ? (a + 200) : c;
		int y = (d == 0) ? b : (d + 200);
		out[x]++;
		in[y]++;
		uf[find(x)] = find(y);		
	}
	for(int i = 1; i <= h; i++){
		if(out[i] > in[i]) return !printf("NO\n");
		if(in[i] > out[i]) has[find(i)] = 1;
	}
	for(int i = 200 + 1; i <= 200 + h; i++){
		if(in[i] > out[i]) return !printf("NO\n");
		if(in[i] < out[i]) has[find(i)] = 1;
	}
	for(int i = 1; i <= h; i++){
		if((in[i] || out[i]) && !has[find(i)]) return !printf("NO\n");
		if((in[200+i] || out[200+i]) && !has[find(200+i)]) return !printf("NO\n");
	}
	printf("YES\n");
	return 0;
}
