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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using pll = pair<ll, ll>;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define Rep(i, m, n) for(ll i = m; i < n; i++)
#define all(vec) vec.begin(), vec.end()
#define tmax(x, y, z) max((x), max((y), (z)))
#define tmin(x, y, z) min((x), min((y), (z)))
#define MINUS(a) memset(a, 0xff, sizeof(a))
#define ZERO(a) memset(a, 0, sizeof(a))
const ll INF = 1LL << 60;
const ll MM = 1000000000; const ll MOD = MM + 7; const ll MMM=9223372036854775807;//2^63 -1
#define ADD(a,b) a = (a + ll(b)) % MOD
#define MUL(a,b) a = (a * ll(b)) % MOD
ll GCD(ll x, ll y){ if(y == 0) return x; else return GCD(y, x % y);}
ll LCM(ll x, ll y){ return x / GCD(x, y) * y;}
template<class T> inline bool chmin(T& a, T b){ if(a > b){ a = b; return true;} return false;}
template<class T> inline bool chmax(T& a, T b){ if(a < b){ a = b; return true;} return false;}
template<typename A, size_t N, typename T> void FILL(A (&array)[N], const T &val){ fill( (T*)array, (T*)(array+N), val );}
using Graph = vector<vector<ll>>;
ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
ll ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
//cout << fixed << setprecision(10);


int main(){
    ll n, m, ans = 0; cin >> n >> m;
    vector<pll> v(m);
    rep(i, m){
        ll a, b; cin >> a >> b; a--, b--;
        v[i] = pll(b, a);
    }
    sort(all(v));
    ll bre = -1;
    rep(i, m){
        if(v[i].second >= bre){
            ans++;
            bre = v[i].first;
        }
    }
    cout << ans << endl;
}