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 <iostream>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<deque>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>

#define rep(i, n) for(int i=0;i<n;i++)

typedef int long long ll;
using namespace std;
typedef pair<int, int> P;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const ll MOD = 1e9 + 7;
static const int MAX = 100;
static const int INF = (1 << 23);

int main() {
    int n,m;
    cin>>n>>m;
    vector<pair<int,int>> v;


    rep(i,n){
        int a,b;
        cin>>a>>b;
        v.push_back({a,b});
    }
    sort(v.begin(),v.end());
    int hi=0;

    ll ans=0;
    int p=0;

    priority_queue<int> q;

    for(int i=1;i<=m;i++){
        while(true){
            if(v[p].first>i||p>=n)break;
            q.push(v[p].second);

            p++;
        }
        if(!q.empty()) {

            ans += q.top();
            q.pop();

        }

    }



    cout<<ans<<endl;


    return 0;
}