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: Wrong Answer)

#include<bits/stdc++.h>
# define  _deb(x) { cerr << "\033[94m"<< "[ " << #x << " = " << (x) << " ]\n\033[0m"; }
using namespace std;
#define int long long
bool cmp(const pair<int,int>&a, const pair<int,int>&b){
	if(a.first != b.first) return a.first > b.first;
	return a.second > b.second;
}
signed main(){
	int n,m;
	cin >> n >> m;
	vector<pair<int , int > > v;
	for(int i=0;i<n;i++){
		int x,y;
		cin >> x >> y;
		v.push_back({y,x});
	}
	sort(v.begin(),v.end(),cmp);
	//for(auto e:v) cout<<e.first<<" " << e.second << endl;
	long long ans = 0;
	for(int i=0;i<v.size() && i<m;i++){
		if(i + v[i].second  <= m){
			ans += v[i].first;
		}
	}
	cout << ans << '\n';
}