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 <memory.h>

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <utility>
#include <vector>

using namespace std;

#define mod 1000000007

int main() {
	int n, m;
	cin >> n >> m;
	vector<pair<int, int> > ab;
	for (int i = 0; i < n; i++) {
		int a, b;
		cin >> a >> b;
		ab.push_back(make_pair(a, b));
	}
	sort(ab.begin(), ab.end());
	long long int ans = 0;
	int cnt = 0;
	priority_queue<int> qu;
	for (int i = 0; i <= m; i++) {
		if (!qu.empty()) {
			ans += qu.top();
			qu.pop();
		}
		while (cnt < n && ab[cnt].first <= i + 1) {
			qu.push(ab[cnt].second);
			cnt++;
		}
	}
	cout << ans << endl;
}