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>

using namespace std;

int n, cnt[1005], in[1005], out[1005], par[1005], ta, tb, tc, td;
bool o[1005];

int find(int a) {
	if (par[a] == a)
		return a;
	return par[a] = find(par[a]);
}

int main() {
	scanf("%d%d", &n, &ta);
	for (int i = 0; i < 1005; i++)
		par[i] = i;
	while (n--) {
		scanf("%d%d%d%d", &ta, &tb, &tc, &td);
		if (tc == 0)
			ta = 500 + ta;
		else
			ta = 500 - tc;
		if (td == 0)
			tb = 500 - tb;
		else
			tb = 500 + td;
		if (find(ta) != find(tb))
			par[par[ta]] = par[tb];
		out[ta]++;
		in[tb]++;
	}
	for (int i = 0; i < 1005; i++)
		if (in[i] * out[i] == 0 || in[i] != out[i])
			o[i] = o[find(i)] = 1;
	for (int i = 0; i < 1005; i++)
		if ((i > 500 && out[i] < in[i]) || (i < 500 && out[i] > in[i]) || !o[find(i)]) {
			printf("NO\n");
			return 0;
		}
	printf("YES\n");
	return 0;
}
