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>

int deg[1234];

int edges[1234][1234];

void dfs(int v, int n) {
  int cur = 0;
  while (true) {
    while (cur <= n && edges[v][cur] == 0) ++cur;
    if (cur > n) break;
    --edges[v][cur];
    dfs(cur, n);
  }
}

int main() {
  int n, h;
  scanf("%d%d", &n, &h);
  for (int i = 0; i < n; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int from = c == 0 ? a : (c + 2 * h);
    int to = d == 0 ? (b + 2 * h) : d;
    deg[from]--;
    deg[to]++;
    edges[from][to]++;
  }
  for (int i = 0; i < 4 * h; i++) {
    if (deg[i] != 0) {
      if ((deg[i] < 0 && i >= 2 * h) || (deg[i] > 0 && i < 2 * h)) {
        puts("NO");
        return 0;
      }
      if (deg[i] < 0) {
        edges[4 * h][i] = -deg[i];
      } else {
        edges[i][4 * h] = deg[i];
      }
    }
  }
  dfs(4 * h, 4 * h);
  for (int i = 0; i <= 4 * h; i++) {
    for (int j = 0; j <= 4 * h; j++) {
      if (edges[i][j] != 0) {
        puts("NO");
        return 0;
      }
    }
  }
  puts("YES");
}
