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 <queue>
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <map>
#include <stdio.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for(int i = (a); i < (int)(b); i++)
#define MAX(a, b) ((ll)(a) > (ll)(b)) ? (a) : (b)
#define MIN(a, b) ((ll)(a) > (ll)(b)) ? (b) : (a)

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vint;
typedef vector<string> vstr;
typedef pair<ll, ll> pint;
unsigned long long inf = (1LL << 62);
ll mod = 1000000007;
pint dx[4] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1}};




ll N, M;

ll Left[100005], Right[100005];

int main() {
	ll a, b;
	cin >> N >> M;


	rep(i, M) {
		cin >> a >> b;

		Left[a] += 1;
		Right[b] += 1;

	}


	stack<ll> s;
	ll ans = 1;
	ll rindex = 0;
	rep(i, N + 1) {
		

		if (Right[i] != 0) {
			rindex = i;
			
		}

		if (Left[i] != 0) {
			if (rindex != 0) {
				ans += 1;
				rindex = 0;
			}

		}

		

	}


	cout << ans << endl;
	
	return 0;
}