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 <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 200005 ;
using namespace std ;

int n, H, e, d[2][N], fa[N], sz[N] ;
bool vis[N] ;

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

int main() {
	scanf("%d%d", &n, &H) ;
	int x1, y1, x2, y2, x, y ;
	rep(i, - H + 1, H) fa[H + i] = H + i ;
	rep(i, 1, n) {
		scanf("%d%d%d%d", &x1, &x2, &y1, &y2) ;
		x = y1 ? H - y1 : H + x1 ;
		y = y2 ? H + y2 : H - x2 ;
		++ d[0][y], ++ d[1][x] ;
		x = getfa(x), y = getfa(y) ;
		if (x != y) fa[x] = y, sz[y] += sz[x] ;
		++ sz[y] ;
	}
	rep(i, 1, H) if (d[0][H + i] > d[1][H + i]) {
		printf("NO\n") ;
		return 0 ;
	}
	rep(i, - H + 1, - 1) if (d[0][H + i] < d[1][H + i]) {
		printf("NO\n") ;
		return 0 ;
	}
	rep(i, - H + 1, H) if (d[0][H + i] != d[1][H + i]) {
		vis[getfa(H + i)] = true ;
	}
	rep(i, - H + 1, H) if (getfa(H + i) == H + i && !vis[H + i] && sz[H + i] > 0) {
		printf("NO\n") ;
		return 0 ;
	}
	printf("YES\n") ;
	return 0 ;
}