Joisino is planning to record N TV programs with recorders.

The TV can receive C channels numbered 1 through C.

The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i.

Here, there will never be more than one program that are broadcast on the same channel at the same time.

When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T).

Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.

Constraints

* 1≤N≤10^5
* 1≤C≤30
* 1≤s_i<t_i≤10^5
* 1≤c_i≤C
* If c_i=c_j and i≠j, either t_i≤s_j or s_i≥t_j.
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N C
s_1 t_1 c_1
:
s_N t_N c_N


Output

When the minimum required number of recorders is x, print the value of x.

Examples

Input

3 2
1 7 2
7 8 1
8 12 1


Output

2


Input

3 4
1 3 2
3 4 4
1 4 3


Output

3


Input

9 4
56 60 4
33 37 2
89 90 3
32 43 1
67 68 3
49 51 3
31 32 3
70 71 1
11 12 3


Output

2
import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int c = sc.nextInt();
    int[][] imos = new int[c][100002];
    for(int i = 0; i < n; i++) {
      int s = sc.nextInt();
      int t = sc.nextInt();
      int u = sc.nextInt();
      imos[u - 1][s]++;
      imos[u - 1][t + 1]--;
    }
    for(int i = 0; i < c; i++) {
      for(int j = 1; j < 100002; j++) {
        imos[i][j] += imos[i][j - 1];
      }
    }
    for(int i = 0; i < c; i++) {
      for(int j = 1; j < 100002; j++) {
        if(imos[i][j] > 0) imos[i][j] = 1; 
      }
    }
    int ans = 0;
    for(int j = 1; j < 100002; j++) {
      int sum = 0;
      for(int i = 0; i < c; i++) {
        sum += imos[i][j];
      }
      ans = Math.max(ans, sum);
    }
    System.out.println(ans);
  }
}