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<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
//#define int long long 
using namespace std;
const int MAXN = 2 * 1e5 + 10, INF = 1e9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
struct Node {
	int x, y;
	bool operator < (const Node &rhs) const {
		return x == rhs.x ? y < rhs.y : x < rhs.x;
	}
}P[MAXN];
main() {
	N = read(); M = read();
	for(int i = 0; i < M; i++) {
		int x = read(), y = read();
		P[i] = (Node) {y, x};
	}
	int ans = 0, last = -1;
	sort(P , P + M);
	for(int i = 0; i < M; i++) {
		int j = i + 1;
		while(j < M && P[j].x == P[i].x) j++;
		if(P[j - 1].y >= last) 
			ans++, last = P[j - 1].x;
		i = j;
	}
	printf("%d", ans);
	return 0;
}