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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <string>
#include <iterator>
#include <map>
#include <set>
#include <iomanip>
#include <vector>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
#define INF 100000000000
#define MAX 200001
#define MOD 1000000007
ll fac[MAX], finv[MAX], inv[MAX];
const int MX = 1000005;

int main(){
    int N, M; 
    cin >> N >> M;
    vector<P> AB(N);
    priority_queue<int> que;
    for(int i = 0; i < N; i++){
        int a, b; 
        cin >> a >> b;
        AB[i] = {a, b};
    }
    sort(AB.begin(), AB.end());
    int idx = 0;
    int day = 1;
    int ans = 0;
    while(day != M + 1){
        int cnt = 0;
        for(int i = idx; i < N; i++){
            if(AB[i].first <= day){
                cnt++;
                que.push(AB[i].second);
            }
        }
        idx += cnt;
        day++;
        if(!que.empty()){
            ans += que.top();
            que.pop();
        }
    }
    cout << ans << endl;
}