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 ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

vector<int> g[401];
int in[401], out[401];
bool vis[401];

bool go(int v){
	if (g[v].empty())return true;
	vis[v] = true;
	bool ok = in[v] != out[v];
	f(i, 0, g[v].size())if (!vis[g[v][i]])ok |= go(g[v][i]);
	return ok;
}

int main(){
	int n, h;
	scanf("%d%d", &n, &h);
	f(i, 0, n){
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		int x, y;
		if (c == 0)x = a; else x = c + 200;
		if (d == 0)y = b + 200; else y = d;
		++in[y];
		++out[x];
		g[x].push_back(y);
		g[y].push_back(x);
	}
	bool ok = true;
	f(i, 0, h + 1)if (in[i] > out[i])ok = false;
	f(i, 201, h + 201)if (out[i] > in[i])ok = false;
	f(i, 0, 401)if (!vis[i])ok &= go(i);
	printf("%s\n", ok ? "YES" : "NO");
}