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>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define EB emplace_back

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;

const int INF = 1e9 + 7;

pi a[100005];

int main()
{
	IO_OP;
	
	int n, m;
	cin >> n >> m;
	for(int i=0;i<n;i++)
		cin >> a[i].S >> a[i].F;
	sort(a, a+n, greater<pi>());
	set<int> s;
	for(int i=0;i<=m;i++)
		s.insert(i);
	int ans = 0;
	for(int i=0;i<n;i++) {
		int l = m - a[i].S;
		if(s.upper_bound(l) == s.begin()) continue;
		int t = *prev(s.upper_bound(l));
		s.erase(t);
		ans += a[i].F;
	}
	cout << ans << endl;

}