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>
#define REP(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n) REP(i,0,n)
typedef long long ll;
using namespace std;

int main() {
    int n, m, ans = 0, cnt;
    cin >> n >> m;
    int a[m], b[m], sum[n+1];
    rep(i, m) {
        scanf("%d %d", &a[i], &b[i]);
    }
    while (true) {
        cnt = 0;
        rep(i, n) sum[i] = 0;
        rep(i, m) {
            if (a[i] >= 0) {
                sum[a[i]-1]++; sum[b[i]-1]--;
                cnt++;
            }
        }
        if (cnt == 0) break;
        ans++;
        rep(i, n) sum[i+1] += sum[i];
        int Max = -1, maxi = -1;
        rep(i, n) {
            if (Max < sum[i]) {
                Max = sum[i];
                maxi = i;
            }
        }
        rep(i, m) {
            if (a[i]-1 <= maxi && maxi < b[i]-1) {
                a[i] = -1;
                b[i] = -1;
            }
        }
    }
    cout << ans << endl;

    return 0;
}