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 <iostream>
#define no return cout << "NO\n", 0
using namespace std;
const int H = 405;
int f[H], d[H];
bool b[H], c[H];
int getf(int x)
{
	return f[x] == x ? x : f[x] = getf(f[x]);
}
int main()
{
	ios::sync_with_stdio(false);
	int n, h;
	cin >> n >> h;
	for (int i = 0; i <= 2 * h; i++)
		f[i] = i;
	while (n--)
	{
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		int x = (c ? -c : a) + h, y = (d ? d : -b) + h;
		f[getf(x)] = getf(y);
		::d[x]++;
		::d[y]--;
		::b[x] = true;
	}
	for (int i = 0; i < h; i++)
		if (d[i] > 0)
			no;
	for (int i = h + 1; i <= 2 * h; i++)
		if (d[i] < 0)
			no;
	for (int i = 0; i <= 2 * h; i++)
	{
		b[getf(i)] |= b[i];
		if (d[i])
			c[f[i]] = true;
	}
	for (int i = 0; i <= 2 * h; i++)
		if (f[i] == i && !c[i] && b[i])
			no;
	cout << "YES\n";
	return 0;
}