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;

vector<int> g[505];
int l_cnt[505], r_cnt[505];
bool used[505];

int print(bool ok) {
        if (ok) cout << "YES" << endl;
        else cout << "NO" << endl;
        return 0;
}

bool dfs(int v) {
        used[v] = true;
        bool res = l_cnt[v] != r_cnt[v];
        for (auto u : g[v]) if (!used[u]) res |= dfs(u);
        return res;
}

int main() {
        int n, h, a, b, c, d;
        cin >> n >> h;
        for (int i = 0; i < n; i ++) {
                cin >> a >> b >> c >> d;
                int lh = (c == 0 ? a : c + 250);
                int rh = (d == 0 ? b + 250 : d);
                l_cnt[lh] ++;
                r_cnt[rh] ++;
                g[lh].push_back(rh);
                g[rh].push_back(lh);
        }
        for (int i = 0; i < h + 1; i ++) if (r_cnt[i] > l_cnt[i]) return print(0);
        for (int i = 251; i < h + 251; i ++) if (l_cnt[i] > r_cnt[i]) return print(0);
        for (int i = 0; i < 501; i ++) if (!used[i]) if (!g[i].empty()) if (!dfs(i)) return print(0);
        return print(1);
}
