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;
#define MAXN	5005
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
int n, m, ind[MAXN], outd[MAXN];
int tot, cnt, home[MAXN];
vector <int> A[MAXN];
bool visited[MAXN];
void work(int pos) {
	if (visited[pos]) return;
	visited[pos] = true;
	home[++tot] = pos;
	cnt += A[pos].size();
	for (unsigned i = 0; i < A[pos].size(); i++)
		work(A[pos][i]);
}
int main() {
	read(n), read(m);
	for (int i = 1; i <= n; i++) {
		int a, b, c, d, inp, outp;
		read(a), read(b), read(c), read(d);
		if (c == 0) inp = a;
		else inp = m + c;
		if (d == 0) outp = m + b;
		else outp = d;
		outd[inp]++; ind[outp]++;
		A[inp].push_back(outp);
		A[outp].push_back(inp);
	}
	for (int i = 1; i <= 2 * m; i++) {
		tot = 0; cnt = 0; work(i);
		if (tot == 0 || cnt == 0) continue;
		bool flg = true;
		for (int i = 1; i <= tot; i++) {
			if (home[i] <= m && ind[home[i]] > outd[home[i]]) {
				printf("NO\n");
				return 0;
			}
			if (home[i] > m && ind[home[i]] < outd[home[i]]) {
				printf("NO\n");
				return 0;
			}
			if (ind[home[i]] != outd[home[i]]) flg = false;
		}
		if (flg) {
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
}