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 <cstring>
const int N = 4e2 + 100;
int f[N];
int ff(int x) {
	return f[x] == x ? x : f[x] = ff(f[x]);
}
int n, h;
int in[N], ou[N], ok[N];

int main() {
	scanf("%d%d", &n, &h);
	for (int i = 1; i <= h + h; ++i)
		f[i] = i;
	for (int i = 1; i <= n; ++i) {
		int a, b, c, d, l, r;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		l = c == 0 ? h + a : c;
		r = d == 0 ? b : h + d;
		++ou[l], ++in[r];
		if (ff(l) != ff(r))
			f[ff(l)] = ff(r);
	}
	for (int i = 1; i <= h; ++i)
		if (ou[i] > in[i])
			return puts("NO"), 0;
	for (int i = h + 1; i <= h + h; ++i)
		if (in[i] > ou[i])
			return puts("NO"), 0;
	for (int i = 1; i <= h + h; ++i)
		if (in[i] != ou[i])
			ok[ff(i)] = true;
	for (int i = 1; i <= h + h; ++i)
		if (ff(i) == i && !ok[i] && in[i] + ou[i] != 0)
			return puts("NO"), 0;
	return puts("YES"), 0;
}