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: Accepted)

#include <bits/stdc++.h>
#define BIT(n) (1LL << (n))
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPB(i, n) for (int i = 0; i < BIT(n); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORA(i, n) for (auto &&i : n)
#define FORS(i, m, n) for (int i = m; i <= n; i++)
#define DEBUG(x) cerr << #x << ": " << x << "\n"
#define ALL(v) v.begin(), v.end()
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
#define epb emplace_back
#define int long long
using namespace std;
typedef long long ll;
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 (a>b) { a=b; return 1; } return 0; }
template < class T > using vec = std::vector< T >;
template < class T > void print(T &x) { cout << x << "\n"; }
const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI = acos(-1.0);
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
signed main() {
  cin.tie(0), ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  int N,M;
  cin >> N>>M;
  vec<vec<int>>ab(N);
  REP(i,M){
    int a,b;
    cin>>a>>b;
    ab[a].pb(b);
  }
  //区間スケジューリング問題として解く。
vec<int>dp(N+1,0);//最大値なので0でよい。[0,i)区間の最大スケジューリング
    REP(i,N){
        //何もしない場合
        chmax(dp[i+1],dp[i]);
        FORA(j,ab[i])chmax(dp[j],dp[i]+1);
    }
    print(dp[N]);
  return 0;
}