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 <iostream>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>
#include <map>
#include <queue>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define REP(i, n) for (int i = 0; i < (n); i++)
#define RREP(i, n) for (int i = (n) - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (a); i > (b); i--)
#define ALL(a) (a).begin(), (a).end()

int main() {
    int N, M;
    pii BA[100000];
    cin >> N >> M;
    REP(i, N) {
        int A, B;
        cin >> A >> B;
        BA[i] = {B, A};
    }
    sort(BA, BA + N, [](const pii& a, const pii& b) { return a.second < b.second; });

    int BA_i = 0;
    ll ans = 0;
    priority_queue<pii> que;
    FOR(i, 1, M + 1) {
        while (BA_i < N && BA[BA_i].second <= i) {
            que.push(BA[BA_i++]);
        }
        if (que.empty()) continue;

        pii t = que.top(); que.pop();
        ans += t.first;
    }
    cout << ans << endl;
    return 0;
}