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)

#include <bits/stdc++.h>
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define REP(i, n) FOR(i, 0, n)
#define all(x) (x).begin(), (x).end()
#define bit(x) (1L << (x))
using ll = long long;
using namespace std;

template<typename T>
vector<T> make_v(size_t a,T b){return vector<T>(a,b);}
 
template<typename... Ts>
auto make_v(size_t a,Ts... ts){
    return vector<decltype(make_v(ts...))>(a,make_v(ts...));
}

int main() {
    int n, m; cin >> n >> m;
    vector<pair<int, int>> p(n);
    REP(i, n) {
        int a, b; cin >> a >> b;
        p[i] = make_pair(a, b);
    }

    sort(p.begin(), p.end(), [](auto x, auto y) {
        if(x.second != y.second)
            return x.second > y.second;
        else
            return x.first < y.first;
    });

    ll ans = 0;
    REP(i, n) {
        if(i + p[i].first <= m)
            ans += p[i].second;
    }

    cout << ans << endl;

    return 0;
}