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>
#define rep(i, l, r) for(int i = (l), i##end = (r);i <= i##end;++i)
using std::cin; using std::cout;
const int maxn = 200200;
typedef long long ll;
const int mod = 998244353;
inline ll pow(ll a,int b,int ans = 1) {
	for(;b;b >>= 1,a = a * a % mod) if(b & 1)
		ans = ans * a % mod;
	return ans;
}
inline ll inverse(int x){ return pow(x, mod - 2); }
int n, h;
int fa[maxn], in[maxn], out[maxn], can[maxn];
inline int find(int x) {
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int main() {
	std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> h;
	rep(i, 1, h + h) fa[i] = i;
	rep(i, 1, n) {
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		int x0 = c ? c + h : a;
		int x1 = d ? d : b + h;
		fa[find(x0)] = find(x1);
		++ out[x0], ++ in[x1];
	}
	rep(i, 1, h) {
		if(out[i] < in[i] || out[i + h] > in[i + h]) {
			cout << "NO" << '\n';
			return 0;
		}
		if(out[i] != in[i]) can[find(i)] = 1;
		if(out[i + h] != in[i + h]) can[find(i + h)] = 1;
	}
	rep(i, 1, h + h) {
		if(in[i] && out[i] && !can[find(i)]) {
			cout << "NO" << '\n';
			return 0;
		}
	}
	cout << "YES" << '\n';
}
