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
//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define F first
#define S second
#define _sz(x) (int)x.size()
#define pb push_back

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int N = 1e5 + 20 , MX = 400 + 20 ;
int n , H , a[N] , b[N] , c[N] , d[N] ;

vector <int> g[MX] , h[MX] ;

bool vis[MX] , c1 , c2 ;

inline void err () {
	cout << "NO\n" ;
	exit(0) ;
}

void dfs (int v) {
	vis[v] = 1 ;
	if (v < MX / 2 && _sz(h[v]) > _sz(g[v])) c1 = 1 ;
	if (v > MX / 2 && _sz(h[v]) < _sz(g[v])) c2 = 1 ;

	for (int u : g[v]) {
		if (!vis[u]) dfs(u) ;
	}
	for (int u : h[v]) {
		if (!vis[u]) dfs(u) ;
	}
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	cin >> n >> H ;
	for (int i = 0 ; i < n ; i ++) {
		cin >> a[i] >> b[i] >> c[i] >> d[i] ;
	
		int l , r ;

		if (c[i]) l = -c[i] ;
		else l = a[i] ;

		if (d[i]) r = d[i] ;
		else r = -b[i] ;
	
		l += MX / 2 , r += MX / 2 ;

		g[l].pb(r) ;
		h[r].pb(l) ;
	}

	for (int i = 0 ; i < MX ; i ++) {
		if (i < MX / 2 && _sz(g[i]) > _sz(h[i])) err() ;
		if (i > MX / 2 && _sz(g[i]) < _sz(h[i])) err() ;
	}

	for (int i = 0 ; i < MX ; i ++) {
		if (_sz(g[i]) + _sz(h[i]) && !vis[i]) {
			c1 = 0 ;
			c2 = 0 ;
			dfs(i) ;

			if (!c1 || !c2) err() ;
		}
	}
	
	cout << "YES\n" ;
}
