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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 405;
int N, H, fa[MAX_N], in[MAX_N], out[MAX_N], ok[MAX_N];
bool vis[MAX_N];

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

int main() {
  scanf("%d%d", &N, &H);
  for (int i = 1; i <= H << 1; ++i) fa[i] = i;
  
  for (int i = 1; i <= N; ++i) {
    int A, B, C, D, u, v;
    scanf("%d%d%d%d", &A, &B, &C, &D);
    if (C == 0) out[A]++, u = A;
    else out[C + H]++, u = C + H;
    
    if (D == 0) in[B + H]++, v = B + H;
    else in[D]++, v = D;

    vis[u] = vis[v] = 1;
    if (find(u) != find(v)) fa[find(u)] = find(v);
  }

  bool flag = 1;
  for (int i = 1; i <= H; ++i)  {
    if (in[i] > out[i]) flag = 0;
    if (in[i + H] < out[i + H]) flag = 0;
    if (in[i] != out[i]) ok[find(i)] = 1;
    if (in[i + H] != out[i + H]) ok[find(i + H)] = 1;
  }
  for (int i = 1; i <= H << 1; ++i)
    if (vis[i] && !ok[find(i)]) flag = 0;
  
  if (flag) puts("YES");
  else puts("NO");
  
  return 0;
}
