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 <iostream>
#include <vector>
using namespace std;
#pragma warning (disable: 4996)

int N, H, A[100009], B[100009], C[100009], D[100009], L[100009], R[100009], deg[100009], col[100009], cnts, cnt; bool used[100009];
vector<int>x[100009];

void dfs1(int pos) {
	if (col[pos] >= 1) return;
	col[pos] = cnts;
	for (int i = 0; i < x[pos].size(); i++) dfs1(x[pos][i]);
}

int main() {
	cin >> N >> H; H++;
	for (int i = 1; i <= N; i++) {
		cin >> A[i] >> B[i] >> C[i] >> D[i];
		if (C[i] == 0) { L[i] = -A[i]; cnt++; }
		else { L[i] = C[i]; }
		if (D[i] == 0) { R[i] = -B[i]; cnt++; }
		else { R[i] = D[i]; }
		R[i] *= -1;
	}
	for (int i = 1; i <= N; i++) {
		int a1 = 0, a2 = 0;
		if (L[i] < 0) a1 = H + (-L[i]); else a1 = L[i];
		if (R[i] < 0) a2 = H + (-R[i]); else a2 = R[i];
		deg[a1]++; deg[a2]--; used[a1] = true; used[a2] = true;
		x[a1].push_back(a2); x[a2].push_back(a1);
	}
	bool OK = true; int V = 0;
	for (int i = 0; i < H; i++) { if (deg[i] > 0) OK = false; }
	for (int i = H; i < 2 * H; i++) { if (deg[i] < 0) OK = false; else V += deg[i]; }

	for (int i = 0; i < 2 * H; i++) {
		if (col[i] == 0 && used[i] == true) { cnts++; dfs1(i); }
	}
	if (cnts > (cnt - N - 1) + 1) OK = false;

	if (OK == true) cout << "YES" << endl;
	else cout << "NO" << endl;
	return 0;
}