Problem:
Score : 400 points

Problem Statement

There are N islands lining up from west to east, connected by N-1 bridges.

The i-th bridge connects the i-th island from the west and the (i+1)-th island from the west.

One day, disputes took place between some islands, and there were M requests from the inhabitants of the islands:

Request i: A dispute took place between the a_i-th island from the west and the b_i-th island from the west. Please make traveling between these islands with bridges impossible.

You decided to remove some bridges to meet all these M requests.

Find the minimum number of bridges that must be removed.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq a_i < b_i \leq N

All pairs (a_i, b_i) are distinct.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the minimum number of bridges that must be removed.

Sample Input 1

5 2
1 4
2 5

Sample Output 1

1

The requests can be met by removing the bridge connecting the second and third islands from the west.

Sample Input 2

9 5
1 8
2 7
3 5
4 6
7 9

Sample Output 2

2

Sample Input 3

5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5

Sample Output 3

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

using namespace std;

typedef long long int lli;
typedef pair<int, int> ii;

template<class T>bool chmax(T &a, const T &b) { if( a < b ) { a = b; return 1; } return 0;}
template<class T>bool chmin(T &a, const T &b) { if( b < a ) { a = b; return 1; } return 0;}

#ifdef LOCAL
	#define eprintf(...) printf(__VA_ARGS__)
#else
	#define eprintf(...) 1
#endif

#define ROUNDUP(a, b) 		( ((a) + ((b) - 1) ) / (b))
#define SET_ZERO(a)	 		( memset((a), 0, sizeof(a)))
#define SET_NUM_ARRAY(a,b)	( memset((a), b, sizeof(a)))
//#define int long long int //dont forget to change int to signed

int main() {
	int N, M;
	cin >> N >> M;
	int bridge[100009];
	for (int i = 0; i < M; i++) {
		int a, b;
		cin >> a >> b;
		bridge[a-1] |= 1;
		bridge[b-2] |= 2;
	}

	int ctr = 0;
	int d = 0;
	for (int i = 0; i < N; i++) {
		if (bridge[i] & 1) {
			d++;
		}
		if (bridge[i] & 2) {
			if (d > 0) {
				d = 0;
				ctr++;
			}
		}
	}
	printf("%d", ctr);
	return 0;
}