Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

The Maximum Number of Customers

$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.

Constraints

$ 1 \leq N \leq 10^5 $

$ 1 \leq T \leq 10^5 $

$ 0 \leq l_i < r_i \leq T $

Input

The input is given in the following format.

$N$ $T$

$l_1$ $r_1$

$l_2$ $r_2$

:

$l_N$ $r_N$

Output

Print the maximum number of persons in a line.

Sample Input 1

6 10
0 2
1 3
2 6
3 8
4 10
5 10

Sample Output 1

4

Sample Input 2

2 2
0 1
1 2

Sample Output 2

1

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#define TMAX 100001
#define THALF 50000

void fill(int *p, int T, int d)
{
	int i;
	for(i=0; i<T; i++){
		*(p+i) = d;
	}
}
void stak(int *p, int head, int end, int d)
{
	int i;
	for(i=head; i<end; i++){
		*(p+i) += d;
	}
}

int main(void){
    int N, T, t[TMAX], entry, exit, d, sw, max=0, i, j;
    scanf("%d%d", &N, &T);
	scanf("%d%d", &entry, &exit);
	sw = (THALF > exit-entry ? 1 : 2);
	switch(sw){
		case 1:
			fill(t, T, 0);
			stak(t, entry, exit, 1);
			for(i=1; i<N; i++){
				scanf("%d%d", &entry, &exit);
				stak(t, entry, exit, 1);
    		}
		    for(j=0; j<T; j++){
		        max = (max < *(t+j) ? *(t+j) : max);
		    }
			break;
		case 2:
			fill(t, T, N);
			stak(t, 0, entry, -1);
			stak(t, exit, T, -1);
			for(i=1; i<N; i++){
				scanf("%d%d", &entry, &exit);
				stak(t, 0, entry, -1);
				stak(t, exit, T, -1);
			}
			for(j=0; j<T; j++){
				max = (max < *(t+j) ? *(t+j) : max);
			}
			break;
		default:
			break;
	}
    printf("%d\n", max);
    return 0;
}