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
// Jigsaw
// * frank_c1
// * 2017 / 11/ 07

#include <bits/stdc++.h>
using namespace std;

const int maxn = (int)(1e5) + 5;
int ai[maxn], bi[maxn], ci[maxn], di[maxn];
int in[maxn], ou[maxn], vi[maxn];
vector<int> nxt[maxn];
int H, cnt, sgn;

void dfs(int x) {
	vi[x] = 1; if (in[x] != ou[x]) sgn = 1; cnt += ou[x] - in[x];
	for (int i = 0; i < (int)nxt[x].size(); ++i) if (!vi[nxt[x][i]]) dfs(nxt[x][i]);
}

int main() {
	int n;
	scanf("%d%d", &n, &H);
	for (int i = 1; i <= n; ++i) {
		scanf("%d%d%d%d", &ai[i], &bi[i], &ci[i], &di[i]);
		int u = 0, v = 0;
		if (!ci[i]) u = ai[i]; else u = ci[i] + H;
		if (!di[i]) v = bi[i] + H; else v = di[i];
		++ou[u]; ++in[v]; nxt[u].push_back(v); nxt[v].push_back(u);
	} int cnt = 0, e = 0;
	for (int i = 1; i <= H; ++i) if (in[i] > ou[i]) return printf("NO\n"), 0; 
	for (int i = H + 1; i <= H + H; ++i) if (in[i] < ou[i]) return printf("NO\n"), 0;
	for (int i = 1; i <= H + H; ++i) if (!vi[i] && in[i] + ou[i]) {
		cnt = sgn = 0; dfs(i); if (cnt || !sgn) return printf("NO\n"), 0;
	}
	return printf("YES\n"), 0;
}