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;
typedef long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second

int n, h;

int f(int d, int u){
	return d?(h+d):u;
}

int g(int d, int u){
	return d?d:(h+u);
}

int out[410];
int in[410];
int par[410];
int chk[410];

int r(int x){return (par[x]==x)?x:(par[x]=r(par[x])); }
void join(int a, int b){ par[r(a)]=r(b); }

void In(){
	scanf("%d%d",&n,&h);
	for(int i=1; i<2*h; ++i) par[i]=i;
	for(int i=1; i<=n; ++i){
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		int x = f(c, c+a);
		int y = g(d, d+b);
		join(x, y);
		out[x]++; in[y]++;
	}
}

bool check(){
	for(int i=1; i<=2*h-1; ++i){
		if(out[i] > in[i]){
			if(i > h) return 0;
			chk[r(i)]=1;
		}
		if(out[i] < in[i]){
			if(i <= h) return 0;
		}
	}
	for(int i=1; i<=2*h-1; ++i) if((out[i] || in[i]) && !chk[r(i)]) return 0;
	return 1;
}

int main()
{
	In();
	puts(check()?"YES":"NO");
	return 0;
}
