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<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
const LL mod=1000000007;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};


int main(){
    int n,m;cin >> n >> m;
    vector<int> a(n),b(n);
    vector<P> d;
    for (int i = 0; i < n; i++) {
        cin >> a[i] >> b[i];
        d.pb(mp(m-a[i],b[i]));
    }
    sort(RALL(d));
    LL ans=0;
    priority_queue<LL> q;
    int k=0;
    for (int i = m; i >= 0; i--) {
        while(k<n&&d[k].first>=i){
            q.push(d[k].second);
            k++;
        }
        if(q.empty()) continue;
        LL p=q.top();q.pop();
        ans+=p;
    }
    cout << ans << endl;
    return 0;
}