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)

#pragma GCC optimize("Ofast")
#pragma GCC target ("avx")
#include <bits/stdc++.h>

constexpr int INF = 2147483647;
constexpr long long int INF_LL = 9223372036854775807;
constexpr int MOD = 1000000007;
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

int main() {
	int N, M;
	cin >> N >> M;
	vector<pair<int, int>> ab(N);
	for (int i = 0; i < N; i++)cin >> ab[i].first >> ab[i].second;
	sort(ab.begin(), ab.end());
	ll ans = 0;
	vector<int> ind(100001, 0);
	for (int i = 1; i <= M; i++) {
		for (int j = ind[i]; j < N; j++) {
			if (i < ab[j].first) {
				ind[i + 1] = j;
				break;
			}
			if (j == N - 1)ind[i + 1] = j+1;
		}
	}

	priority_queue<int> q;
	for (int i = M-1; i >= 0; i--) {
		for (int j = ind[M - i]; j < ind[M - i + 1]; j++) {
			q.push(ab[j].second);
		}
		if (!q.empty()) {
			ans += q.top(); q.pop();
		}
	}

	cout << ans << endl;
}