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 <iostream>
#include <vector>
#include <limits.h>
#include <algorithm>
#include <string>
#include <math.h>
#include <limits.h>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <random>
#include <functional>
#include <stack>
#include <iomanip>
using namespace std;

template<class T>
void in(vector<T>& V) {
	for (auto itr = V.begin(); itr != V.end(); itr++)cin >> *itr;
	return;
}

template<class T>
void out(vector<T>& V) {
	for (T& itr : V)cout << itr << endl;
	return;
}

typedef pair<int, int> pii;
vector<pii> A;

int N, M;

int main() {
	cin >> N >> M;
	A.resize(N);
	for (int n = 0; n < N; n++) {
		int a, b;
		cin >> a >> b;
		A[n] = pii(a, b);
	}
	sort(A.begin(), A.end(), [](pii a,pii b) {
		if (a.second != b.second)return a.second > b.second;
		else return a.first > b.first;
	});
	long long ans = 0;
	int l = 0;
	for (int m = 0; m < M;m++) {
		if (m + A[l].first <= M)ans += A[l].second;
		l++;
	}
	cout << ans << endl;
	return 0;
}