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<bits/stdc++.h>
using namespace std;
int n,m,s=1,num;
struct A{
	int p,t;
}a[100010];
bool cmp(const A& a,const A& b) {
	return a.t<b.t;
}
priority_queue<int> ans;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i) scanf("%d%d",&a[i].t,&a[i].p);
	sort(a+1,a+n+1,cmp);
	for(int i=m;i>=1;--i) {
		while(a[s].t+i-1<=m&&s<=n) ans.push(a[s].p),s++;
		if(!ans.empty()) num+=ans.top(),ans.pop();
	}
	printf("%d",num);
	return 0;
}