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 ("O3")
#include <bits/stdc++.h>
#define ll long long
#define cst first
#define dur second
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e5 + 5, M = 2 * N + 5;

int n, m;
int main(){
    scanf("%d%d", &n, &m);
    vector<pair<int, int> > v;
    for(int i = 0 ; i < n ; ++i){
        int a, b;
        scanf("%d%d", &a, &b);
        if(a <= m)
            v.emplace_back(b, -a);
    }
    sort(v.rbegin(), v.rend());
    ll ans = 0;
    int day = m;
    vector<pair<int, int> > m3lsh;
    for(auto p : v){
        if(day == 0)break;
        m3lsh.emplace_back(-p.dur, p.cst);
        --day;
    }
    sort(m3lsh.rbegin(), m3lsh.rend());
    day = m;
    for(auto p : m3lsh){
        if(p.first <= day)
            ans += (1LL * p.second);
        else continue;
        --day;
    }
    printf("%lld\n", ans);
}