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;
#define ll long long
const int N=1e6+444;
ll ans;
struct re {
	int day;
	int val;
} a[N];
int n,ddl;
int t[N],hr,sum[N];
bool cmp(re a,re b) {
	return a.val>b.val||(a.val==b.val&&a.day<b.day);
}
int main() {
	ios::sync_with_stdio(0);
//	freopen("p.in","r",stdin);
//	freopen("p.out","w",stdout);
	cin>>n>>ddl;
	ddl++;
	for(int i=1; i<=n; i++)cin>>a[i].day>>a[i].val;
	sort(a+1,a+1+n,cmp);
	for(int i=1; i<=n; i++) {
		if(ddl==hr+1)break;
		int ed=ddl-a[i].day;
		if(sum[ed]==ed)break;
		//cout<<ed<<endl;
		for(int j=ed; j>=1; j--)
			if(!t[j]) {
				for(int k=j;k<=ddl;k++)sum[k]++;
				t[j]=1;
				hr++;
				ans+=a[i].val;
				break;
			}
	}
	cout<<ans;
	return 0;
}