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
# include <cstdio>
int main()
{
	int N , C ;
	int s [100000] , t [100000] , c [100000];
	int sm [200002];
	int tt [200002];
	
	scanf("%d %d" ,&N ,& C );
	for(int i =0;i < N ;i++)	scanf("%d %d %d" ,& s[i] ,&t[i] ,&c[i]);
	for(int i =1;i <= C ;i++)
	{	
		for(int j =0; j <200002; j++) tt[j]=0;
		for(int j =0; j < N ; j++) if (c[j]== i) tt[s[j]*2-1]++, tt[t[j]*2] --;
		for(int j =1; j <200002; j++) tt[j]+= tt[j-1];
		for(int j =0; j <200002; j++) if(tt[j] >0) sm[j]++;
	}
	int mx =0;
	for (int j =0;j <200002;j++) if( mx < sm[j]) mx = sm[j];
	printf ("%d\n" ,mx );

	return 0;
}