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package org.psStrategies;

import java.util.Arrays;

import java.util.Scanner;

public class test3 {

public static void main(String[] args) {

// ---------

// 방법 1.

// ---------

Scanner sc = new Scanner(System.in);

int n = sc.nextInt();

int[][] arr = new int[n][2];

for (int i = 0; i < n; i++) {

arr[i][0] = sc.nextInt(); // start

arr[i][1] = sc.nextInt(); // end

}

Arrays.sort(arr, (o1, o2) -> {

if (o1[1] == o2[1])

return Integer.compare(o1[0], o2[0]);

else

return Integer.compare(o1[1], o2[1]);

});

int max = 1;

for (int i = 0; i < n; i++) {

int end = arr[i][1];

int answer = 1;

for (int j = i + 1; j < n; j++) {

int nextstart = arr[j][0];

if (nextstart < end) {

answer++;

}

}

max = Math.max(max, answer);

}

System.out.println(max);

// ---------

// 방법 2. 방문하고 있는 시간에 부스사용하고있으면 +1, 가장 많이 사용한 부스만큼 있어야 대기하지 않고 작업끝낼수있다.

// ---------

int[] arr2 = new int[161]; // 제일 늦게 끝나도 160을넘을수없다

int max2 = -1;

for (int i = 0; i < n; i++) {

int start = sc.nextInt();

int end = sc.nextInt();

for (int j = start; j < end; j++) {

arr2[j]++;

if(max2 < arr2[j])

max2 = arr2[j];

}

}

}

}