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"
#include<assert.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
const int inf = 1e17;
const int mod = 1000000007;
typedef pair<int, int> P;
typedef pair<int, P> PP;
vector<int>G[405];
int in[405];
int a[100005], b[100005], c[100005], d[100005];
int vst[405];
void dfs(int now) {
	for(int e:G[now]) {
		if (vst[e])continue;
		vst[e] = 1;
		dfs(e);
	}
}
signed main() {
	int n, h; cin >> n >> h;
	rep(i, n) {
		cin >> a[i] >> b[i] >> c[i] >> d[i];
		int x, y;
		if (c[i] == 0)x = a[i];
		else x = c[i] + h;
		if (d[i] == 0)y = b[i] + h;
		else y = d[i];
		G[x].push_back(y);
		in[y]++;
	}
	int cnt = 0;
	rep(i, h + 1) {
		int z = G[i].size();
		if (z-in[i]>=1) {
			cnt+=z-in[i];
			G[0].push_back(i);
			in[i]++;
		}
		else if (z != in[i]) {
			puts("NO"); return 0;
		}
	}
	for (int i = h + 1; i <= 2 * h; i++) {
		int z = G[i].size();
		if (in[i]-z>=1) {
			cnt-=in[i]-z;
			G[i].push_back(0);
			in[0]++;
		}
		else if (z != in[i]) {
			puts("NO"); return 0;
		}
	}


	if (cnt!=0) {
		puts("NO");
	}
	else {
		vst[0] = 1;
		dfs(0);
		rep(i, 2 * h + 1) {
			if (vst[i] == 0 && (in[i]||G[i].size())) {
				puts("NO"); return 0;
			}
		}

		puts("YES");
	}
}