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 = 409;
int in[maxn], out[maxn], pa[maxn], flag[maxn];
int n, H, a, b, c, d;

int getpa(int x) {
	return pa[x] == x ? x : pa[x] = getpa(pa[x]);
}

int main() {
	scanf("%d%d", &n, &H);
	for (int i=1; i<=2*H; i++) pa[i] = i;
	for (int i=1; i<=n; i++) {
		scanf("%d%d%d%d", &a, &b, &c, &d);
		if (c) a = c + H; 
		if (!d) b = b + H; else b = d;
		out[a]++; in[b]++;
		pa[getpa(a)] = getpa(b);
	}
	for (int i=1; i<=H; i++)
		if (out[i] < in[i]) { puts("NO"); return 0; }
	for (int i=H+1; i<=2*H; i++)
		if (in[i] < out[i]) { puts("NO"); return 0; }
	for (int i=1; i<=2*H; i++)
		if (in[i] != out[i]) flag[getpa(i)] = 1;
	for (int i=1; i<=2*H; i++)
		if (pa[i] == i && !flag[i] && in[i] && out[i]) { puts("NO"); return 0; }
	puts("YES");
	return 0;
}