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.



<image>

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;
const int N = 1e5 + 10;
int n, h, a, b, c, d, in[500], out[500], f[N], ok[N];
int find(int x) {
    if(x == f[x]) return x;
    return f[x] = find(f[x]);
}
int main() {
    scanf("%d %d", &n, &h);
    for(int i = 1; i <= n; ++i) f[i] = i;
    for(int i = 1; i <= n; ++i) {
        scanf("%d %d %d %d", &a, &b, &c, &d);
        int x = h, y = h;
        if(!d) x -= b; else x += d;
        if(!c) y += a; else y -= c;
        ++out[x]; ++in[y];
        int fx = find(x), fy = find(y);
        if(fx != fy) f[fx] = fy;
    }
    for(int i = 1; i <= h + h; ++i) 
        if(out[i] != in[i] || (!out[i] && !in[i])) ok[find(i)] = 1;
    int ans = 1;
    for(int i = 1; i <= h; ++i) if(in[i] > out[i]) ans = 0;
    for(int i = h + 1; i <= h + h; ++i) if(out[i] > in[i]) ans = 0;
    for(int i = 1; i <= h + h; ++i) if(f[i] == i && !ok[i]) ans = 0;
    printf(ans ? "YES\n" : "NO\n");
    return 0;
}
