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 = 2 * (200 + 10);

int n, H;
vector<int> adj[MAXN], rev[MAXN], vec;
int in[MAXN], out[MAXN];
bool mark[MAXN];

void dfs(int v)
{
	vec.push_back(v);
	mark[v] = true;
	for (auto i : adj[v])
		if (!mark[i])
			dfs(i);
	return;
}


int main()
{
	ios::sync_with_stdio(0);
	cin >> n >> H;
	for (int i = 0; i < n; i++)
	{
		int a, b, c, d, v = 0, u = 0;
		cin >> a >> b >> c >> d;
		if (c)
			v += H + 1 + c;
		else
			v += a;
		if (d)
			u += d;
		else
			u += b + H + 1;
		adj[v].push_back(u);
		adj[u].push_back(v);
		in[u]++;
		out[v]++;
	}
	for (int i = 0; i < H + 1; i++)
		if (out[i] < in[i])
			return cout << "NO" << endl, 0;
	for (int i = H + 1; i < ((H + 1) << 1); i++)
		if (in[i] < out[i])
			return cout << "NO" << endl, 0;
	for (int i = 0; i < MAXN; i++)
	{
		if (!mark[i] && (in[i] + out[i]))
		{
			dfs(i);
			bool can = false;
			for (auto me : vec)
				if (in[me] != out[me])
					can = true;
			if (!can)
				return cout << "NO" << endl, 0;
			vec.clear();
		}
	}
	cout << "YES" << endl;
	return 0;
}