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

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MH = 410;

int N, H;
vi g[MH];
int in[MH], out[MH];
bool vis[MH];

bool dfs(int v) {
	vis[v] = 1;
	bool f = 0;
	if (in[v] != out[v]) f = 1;
	for (int to : g[v]) if (!vis[to]) {
		if (dfs(to)) {
			f = 1;
		}
	}
	return f;
}

int main() {
	scanf("%d %d", &N, &H);
	rep(i, N) {
		int a, b, c, d;
		scanf("%d %d %d %d", &a, &b, &c, &d);
		int l = (c == 0 ? a : -c) + H;
		int r = (d == 0 ? -b : d) + H;
		g[l].pb(r); g[r].pb(l);
		++in[r]; ++out[l];
	}
	for (int i = 1; i <= H; ++i) {
		if (in[i + H] > out[i + H]) {
			puts("NO");
			return 0;
		}
		if (in[H - i] < out[H - i]) {
			puts("NO");
			return 0;
		}
	}

	bool ok = 1;
	rep(i, MH) {
		if (in[i] + out[i] > 0 && !vis[i]) {
			if (!dfs(i)) {
				ok = 0;
			}
		}
	}
	puts(ok ? "YES" : "NO");
	return 0;
}