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>

const int N = 432;

int n, H, V;
int p[N], in[N], out[N];
bool dicyc[N];

int ancestor(int x) {return p[x] == x ? x : (p[x] = ancestor(p[x]));}

int main() {
	int i, Lu, Ld, Ru, Rd, L, R;
	scanf("%d%d", &n, &H), V = 2 * H;
	std::iota(p, p + V, 0);
	for (i = 1; i <= n; ++i)
		scanf("%d%d%d%d", &Lu, &Ru, &Ld, &Rd),
		++out[L = (Ld ? Ld + H : Lu) - 1], ++in[R = (Rd ? Rd : Ru + H) - 1],
		p[ancestor(L)] = ancestor(R);
	for (i = 0; i < H && in[i] <= out[i] && in[i + H] >= out[i + H]; ++i);
	if (i != H) return puts("NO"), 0;
	for (i = 0; i < V; ++i) dicyc[ancestor(i)] |= in[i] != out[i] || !in[i];
	for (i = 0; i < V && (p[i] != i || dicyc[i]); ++i);
	return puts(i == V ? "YES" : "NO"), 0;
}