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 <vector>

const int MH = 405;

int N, H, deg[MH];
std::vector<int> G[MH];

int ok, vis[MH];
void DFS(int u) {
	vis[u] = 1;
	if (deg[u]) ok = 1;
	for (int v : G[u]) if (!vis[v]) DFS(v);
}

int main() {
	scanf("%d%d", &N, &H);
	for (int i = 1; i <= N; ++i) {
		int a, b, c, d, x, y;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		if (!c) x = a;
		else x = c + H;
		if (!d) y = b + H;
		else y = d;
		++deg[x], --deg[y];
		G[x].push_back(y), G[y].push_back(x);
	}
	for (int i = 1; i <= H; ++i)
		if (deg[i] < 0 || deg[i + H] > 0) return puts("NO"), 0;
	for (int i = 1; i <= 2 * H; ++i) if (!G[i].empty() && !vis[i]) {
		ok = 0, DFS(i);
		if (!ok) return puts("NO"), 0;
	}
	puts("YES");
	return 0;
}