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;
int n, h;
const int N = 2000;
int fa[N], ok[N], cnt[N], vis[N];
int get_f(int t) {return fa[t] = (fa[t] == t? t: get_f(fa[t]));}
int main()
{
    cin >> n >> h;
    for (int i = 1; i <= h * 2; ++ i) fa[i] = i;
    for (int i = 1; i <= n; ++ i)
    {
        int a, b, c, d, l, r;
        cin >> a >> b >> c >> d;
        if (c == 0) l = a; else l = c + h;
        if (d == 0) r = b + h; else r = d;
        // cerr << l << " " << r << endl;
        cnt[l] ++; cnt[r] --;
        vis[l] = vis[r] = 1;
        if (get_f(l) != get_f(r)) fa[get_f(l)] = get_f(r);
    }
    for (int i = 1; i <= h; ++ i) if (cnt[i] < 0) return puts("NO"), 0;
    for (int i = 1; i <= h; ++ i) if (cnt[i + h] > 0) return puts("NO"), 0;
    for (int i = 1; i <= h * 2; ++ i) if (cnt[i] != 0) ok[get_f(i)] = 1;
    for (int i = 1; i <= h * 2; ++ i) if (vis[i]) if (!ok[get_f(i)]) return puts("NO"), 0;
    puts("YES");
}
