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 = 405;
int n, m, f[N], delta[N];
bool mark[N], ok[N];

int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }

int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m;
	for (int i = 0; i <= m * 2; ++i) f[i] = i;
	for (int i = 0; i < n; ++i) {
		int a, b, c, d; std::cin >> a >> b >> c >> d;
		int L = (c ?: -a) + m, R = (d ? -d : b) + m;
		mark[L] = mark[R] = 1, f[find(L)] = find(R);
		++delta[L], --delta[R];
	}
	for (int i = 0; i <= m * 2; ++i) if (mark[i])
		if (delta[i] > 0 && i > m || delta[i] < 0 && i < m)
			return std::cout << "NO\n", 0;
	for (int i = 0; i <= m * 2; ++i) ok[find(i)] |= delta[i] != 0;
	for (int i = 0; i <= m * 2; ++i)
		if (mark[i] && i == find(i) && !ok[i])
			return std::cout << "NO\n", 0;
	std::cout << "YES\n";
	return 0;
}