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 <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long

#define maxN 200011
#define maxH 555

int n, h, i;
int a, b, c, d, l, r;
vector<int> list[maxH];
int in[maxH], out[maxH];
bool us[maxN], any;

void no_sol() {
    printf("NO");
    exit(0);
}

void dfs(int node) {
    us[node] = true;
    if (in[node] != out[node]) any = true;
    if (in[node] == 0 && out[node] == 0) any = true;
    for (auto to : list[node])
        if (!us[to])
            dfs(to);
}

int main()
{
 //   freopen("test.in","r",stdin);

    scanf("%d%d", &n, &h);
    for (i = 1; i <= n; i++) {
        scanf("%d%d%d%d", &a, &b, &c, &d);
        if (c == 0) l = a;
        else        l = -c;

        if (d == 0) r = -b;
        else        r = d;

        l += h;
        r += h;
        list[l].pb(r);
        list[r].pb(l);
        in[r]++;
        out[l]++;
    }

    for (i = 1; i <= h; i++)
        if (out[i + h] < in[i + h])
            no_sol();

    for (i = 1; i <= h; i++)
        if (out[-i + h] > in[-i + h])
            no_sol();

    for (i = 1; i <= h; i++) {
        if (!us[i + h]) {
            any = false;
            dfs(i + h);
            if (!any) no_sol();
        }

        if (!us[-i + h]) {
            any = false;
            dfs(-i + h);
            if (!any) no_sol();
        }
    }


    printf("YES");

    return 0;
}
