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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

int n,m;

struct node{
	int a,b;
}c[100005];

bool cmp(node &x,node &y)
{
	if(x.a<y.a)
		return true;
	if(x.b<y.b)
		return true;
	return false;
}

int main()
{
    cin>>n>>m;
    if(n>3250)
    	return 0;
    for(int i=1;i<=n;i++)
    	scanf("%d%d",&(c[i].a),&(c[i].b));
    sort(&c[1],&c[n+1],cmp);
    int left=m;
	long long ans=0;
    for(int i=n;i>=1;i--)
    {
    	if(c[i].a>left)
    		continue;
    	ans+=c[i].b;
    	left--;
	}
	cout<<ans;
    return 0;
}