Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
struct jb
{
	int a,b;
	friend bool operator <(jb x,jb y)
	{
		return x.a>y.a;
	};
}g[100010];
int n,m,k,l,ans=0,f[100010];
set<int> s;
_Rb_tree_const_iterator<int> p;
int mx(int x,int y)
{
	return x>y?x:y;
}
int mn(int x,int y)
{
	return x<y?x:y;
}
void read(int &x)
{
	char c;
	x=0;
	while(c>'9' || c<'0') c=getchar();
	while(c>='0' && c<='9')
	{
		x=x*10+c-'0';
		 c=getchar();
	}
}
int main()
{
	//freopen("ted.in","r",stdin);
	read(n);
	read(m);
	for(int i=0;i<n;i++)
	{
		read(g[i].b);
		read(g[i].a);
	}
	sort(g,g+n);
	for(int i=0;i<=m;i++) s.insert(i);
	for(int i=0;i<n;i++)
	{
		g[i].b=m-g[i].b;
		p=s.lower_bound(g[i].b);
		if(p!=s.begin() && (*p)>g[i].b) p--;
		if((*p)<=g[i].b)
		{
			ans+=g[i].a;
			s.erase(*p);
		}
	}
	printf("%d\n",ans);
	return 0;
}