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<stdio.h>
#include<set>
#include<vector>
#include<algorithm>

using namespace std;

typedef pair<int,int> pii;
typedef long long ll;

const int MX = 100005;

vector<pii> G[MX];
int ind[MX];
int N, H;

int f(int y){ return y <= H ? y + H : y - H; }

int vst[MX];
void dfs(int s){
	for(pii c : G[s]){
		if( vst[c.second] ) continue;
		vst[c.second] = 1;
		dfs(c.first);
	}
}

struct UF{
	int t[MX];
	int find(int x){ return t[x] ? t[x] = find(t[x]) : x; }
	int merge(int a, int b){
		a = find(a), b = find(b);
		return a == b ? 0 : (t[a] = b, 1);
	}
} uf;

int main()
{
	scanf("%d%d", &N, &H);
	for(int i = 1; i <= N; i++){
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		int x, y;
		if( c == 0 ) x = a + H;
		else x = c;
		if( d == 0 ) y = b + H;
		else y = d;
		G[x].emplace_back(f(y), i);
		uf.merge(x, f(y));
		ind[f(y)]++;
	}
	int s = -1, cnt = 0, e = -1;
	for(int i = H+1; i <= 2*H; i++){
		if( G[i].size() < ind[i] ) return !printf("NO\n");
	}
	for(int i = 1; i <= H; i++){
		if( G[i].size() > ind[i] ) return !printf("NO\n");
	}
	set<int> X;
	for(int i = 1; i <= 2*H; i++) X.insert(uf.find(i));
	for(int i = 1; i <= 2*H; i++){
		if( G[i].size() != ind[i] || G[i].size() == 0|| G[i].size() == 0) X.erase(uf.find(i));
	}
	if( X.size() ) return !printf("NO\n");
	printf("YES\n");
}
