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"
#include <string>
using namespace std;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ll long long int
#define int ll
//#define slld(t) scanf("%lld",&t)
//#define sd(t) scanf("%d",&t)
//#define pd(t) printf("%d\n",t)
//#define plld(t) printf("%lld\n",t)
#define vec vector < int >
#define vecp vector < pair < int , int > >
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define debug(x)   cerr<<#x<<": "<<x<<endl
#define pii pair < int , int >
#define pll pair < ll , ll > 
#define debug(x)   cerr<<#x<<": "<<x<<endl
// #define mod  1000000007
#define N 100005
// #define MAX 200005
#define mod1 1000000007
#define mod2 998244353
#define bitcnt __builtin_popcount
#define PI acos(-1)
//#define endl "\n"
// const int inf = (int)1e18 + 10;
const int mod = 1e9 + 7;

const int inf = 1e15;
const int MAX = 1000000;

int a[100005] , b[100005];
vecp v;
bool cmp(pii a , pii b){
  if(a.ff == b.ff)
    return (a.ss > b.ss);
  else return(a.ff > b.ff);
}
signed main(){
  fast;
 
 int n , m;
   cin >> n >> m;
   for(int i = 0 ; i < n ; i++)
      cin >> a[i] >> b[i] , v.pb(mp((b[i] + a[i]) , a[i]));



    // sort(v.begin() , v.end() , cmp);
    sort(v.rbegin() , v.rend());
    int left = m , ans = 0;
    for(int i = 0 ; i < n ; i++){
      // cout << v[i].ff <<"  " <<  v[i].ss<<endl;
          int x = v[i].ff - v[i].ss;
          int y = v[i].ss;

          if(y <= left)
            ans += x , left--;
          if(left == 0) break;
    }

    cout << ans << endl;




// 4  1
// 3  2
// 3  1
// 2  1
// 1  2









}