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<algorithm>
#include<vector>
using namespace std;

bool vis[10010];
vector<int> g[10010];
int ind[10010], outd[10010], fa[10010], p[10010];

int Ask(int x) {
	return x == fa[x] ? x : fa[x] = Ask(fa[x]);
}

int main() {
	int n, h;
	scanf("%d%d", &n, &h);
	for(int i = 0; i <= h + h; i++)
		fa[i] = i;
	for(int i = 1; i <= n; i++) {
		int a, b, c, d, p, q;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		p = !c ? a : -c;
		q = !d ? -b : d;
		int fp = Ask(p + h), fq = Ask(q + h);
		if(fp != fq) 
			fa[fp] = fq;
		ind[q + h]++;
		outd[p + h]++;
	}
	for(int i = 0; i <= h + h; i++) {
		if(i < h && outd[i] > ind[i]) {
			printf("NO\n");
			return 0;
		}
		if(i > h && outd[i] < ind[i]) {
			printf("NO\n");
			return 0;
		}
	}
	for(int i = 0; i <= h + h; i++) {
		if(!ind[i] && !outd[i]) continue;
		int fx = Ask(i);
		p[fx] |= (ind[i] != outd[i]);
	}
	for(int i = 0; i <= h + h; i++) {
		if(!ind[i] && !outd[i]) continue;
		if(Ask(i) != i) continue;
		if(!p[i]) {
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
}