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: Time Limit Exceeded)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<bitset>
#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<tuple>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<stack>
#include<numeric>
#include<functional>

#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define REP(i,n) for(int i = 0;i < n;i++)
#define MEM(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;

int main() {
	IOS;
	vector<P> v;
	int N, M;
	bool br[114514];
	MEM(br, false);
	int ans = 0;
	cin >> N >> M;
	REP(i,M){
		int x, y;
		cin >> x >> y;
		v.push_back(P(y-1, x-1));
	}
	sort(v.begin(), v.end());
	REP(i, M) {
		bool tmp = true;
		for (int j = v[i].first - 1; j >= v[i].second; j--) {
			if (br[j])tmp = false;
		}
		if (tmp) {
			br[v[i].first - 1] = true;
			ans++;
		}
	}

	cout << ans << endl;
}