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 std::cerr;
using std::endl;

const int N = 1e5 + 233;

int n, h, A[N], B[N], C[N], D[N];
int out[N], in[N], fa[N], vis[N];

int find(int x) {
  return x == fa[x] ? x : fa[x] = find(fa[x]);
}

inline bool check() {
  for (int i = 0; i < h; ++i)
    if (in[i] > out[i]) return 0;
  for (int i = 1; i <= h; ++i)
    if (in[i + h] < out[i + h]) return 0;
  for (int i = -h; i <= h; ++i)
    if ((!in[i + h] && !out[i + h]) || (in[i + h] != out[i + h]))
      vis[find(i + h)] = 1;
  for (int i = -h; i <= h; ++i)
    if (!vis[find(i + h)]) return 0;
  return 1;
}

int main() {
  scanf("%d %d", &n, &h);
  for (int i = 0; i <= h * 2 + 5; ++i)
    fa[i] = i;
  for (int i = 1; i <= n; ++i) {
    scanf("%d %d %d %d", A + i, B + i, C + i, D + i);
    int x = C[i] == 0 ? h - A[i] : h + C[i];
    int y = D[i] == 0 ? h + B[i] : h - D[i];
    ++out[x], ++in[y];
    x = find(x), y = find(y);
    if (x != y) fa[x] = y;
  }
  puts(check() ? "YES" : "NO");
  return 0;
}

