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.



<image>

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>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

const int N = 410;

int indeg[N], outdeg[N], fa[N];

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

int n, H;

int main(){
	scanf("%d%d", &n, &H);

	For(i, 1, 2 * H) fa[i] = i;
	For(i, 1, n){
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		int u = !c ? a : H + c, v = !d ? H + b : d;
		fa[find(u)] = find(v);
		++outdeg[u], ++indeg[v];
	}

	bool flag = true;
	For(i, 1, 2 * H) if(i == fa[i] && (indeg[i] || outdeg[i])){
		bool differ = false;
		For(j, 1, H) if(find(j) == i){
			if(outdeg[j] < indeg[j]) flag = false;
			if(outdeg[j] ^ indeg[j]) differ = true;
		}

		For(j, H + 1, 2 * H) if(find(j) == i){
			if(indeg[j] < outdeg[j]) flag = false;
			if(indeg[j] ^ outdeg[j]) differ = true;
		}

		if(!flag || !differ){
			flag = false;
			break;
		}
	}

	puts(flag ? "YES" : "NO");

	return 0;
}