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 <cstdlib>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <tuple>
#include <math.h>
#include <deque>
#include <stack>

using namespace std;

typedef long long ll;
struct data{
    ll d;
    ll s;
};


bool cmpd(const data &a,const data &b)
{
    return a.d > b.d;
}

bool cmps(const data &a,const data &b)
{
    return a.s > b.s;
}

int main(void){
    ll n,m;
    cin >> n >> m;
    vector<data> datas(n);
    for(ll i = 0;i < n;i++){
        data d;
        cin >> d.d >> d.s;
        datas[i] = d;
    }
    sort(datas.begin(), datas.end(), cmpd);
    sort(datas.begin(), datas.end(), cmps);
    ll res = 0;
    for(ll i = 0;i <= m;i++){
        if(i + datas[i].d <= m && datas[i].s != -1){
            res += datas[i].s;
            datas[i].d = 0;
            datas[i].s = -1;
        }
    }
    cout << res << endl;
    return 0;
}