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;
typedef long long ll;



int main() {
#ifdef HOME
    freopen("in", "r", stdin);
#endif

    int n, h;
    while (scanf("%d%d", &n, &h) == 2) {
        vector<int> a(n), b(n), c(n), d(n);
        for (int i = 0; i < n; i++) {
            scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
        }

        vector<int> p(2 * h);
        iota(p.begin(), p.end(), 0);
        function<int(int)> get = [&](int x) {
            return x == p[x] ? x : p[x] = get(p[x]);
        };
        auto uni = [&](int u, int v) {
            p[get(v)] = get(u);
        };
        vector<int> deg(2 * h);
        vector<char> ok(2 * h, 1);
        for (int i = 0; i < n; i++) {
            int u = c[i] == 0 ? a[i]-1 : h + c[i]-1;
            int v = d[i] == 0 ? h+b[i]-1 : d[i]-1;
            ok[u] = ok[v] = 0;
            uni(u, v);
            deg[u]++;
            deg[v]--;
        }
        bool f = 1;
        for (int i = 0; i < h; i++) {
            f &= deg[i] >= 0;
            f &= deg[i + h] <= 0;
        }
        for (int i = 0; i < 2 * h; i++) if (deg[i] != 0) ok[get(i)] = 1;
        for (int i = 0; i < 2 * h; i++) if (p[i] == i && !ok[i]) f = 0;
        string ans = f ? "YES" : "NO";
        cout << ans << endl;
    }

#ifdef HOME
    cerr << clock() / (double)CLOCKS_PER_SEC << endl;
#endif
    return 0;
}
