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<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>
#include<complex>

using namespace std;

#define ll long long
#define ld long double
#define EPS 1e-9
#define INF 1e9
#define LINF (ll)INF*INF
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(int i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

#define int ll //!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef pair<int,pii> pip;
typedef vector<pii> vp;
typedef vector<vi> vvi;

int gcd(int a, int b){if(b==0) return a;return gcd(b,a%b);}
int lcm(int a, int b){return a/gcd(a,b)*b;}


signed main(void) {
    int n,m;
    cin >> n >> m;
    vvi v(m);
    rep(i,n){
        int a,b;
        cin >> a >> b;
        if(m-a >= 0 && m-a < m)v[m-a].push_back(b);
    }
    int ans = 0;
    priority_queue<int> q;
    rep(i,m){
        int now = m-i-1;
        rep(j,v[now].size()){
            //cout << "!" << v[now][j] << endl;
            q.push(v[now][j]);
        }
        if(q.size()){
            ans += q.top();
            q.pop();
        }
    }
    cout << ans << endl;
}