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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<utility>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#include<string>
#define rep(i,n,m) for(int i=(n);i<(int)(m);i++)
#define reps(i,n,m) for(int i=(n);i<=(int)(m);i++)
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define lb lower_bound
#define ub upper_bound
#define LB(a,x) lb(all(a), x) - a.begin()
#define UB(a,x) ub(all(a), x) - a.begin()
#define printfdouble(x) printf("%.12f\n",(x))
#define chartoint(c) (int)((c) - '0')
#define chartoll(c) (long long)((c) - '0')
#define MOD 1000000007
#define itn int
#define enld endl
#define ednl endl
#define icn cin
#define Endl endl
#define stirng string
using namespace std;
typedef long long ll;
const double pi = 3.141592653589793;
using Graph = vector<vector<int>>;
template<class T> bool chmax(T &a, const T &b){if(a<b){a=b; return 1;} return 0;}
template<class T> bool chmin(T &a, const T &b){if(b<a){a=b; return 1;} return 0;}
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}


int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    int N,M; cin >> N >> M;
    vector<int> v[100100];
    rep(i,0,N){
        int a,b; cin >> a >> b;
        v[a].push_back(b);
    }
    rep(i,0,N) sort(v[i].begin(), v[i].end());
    priority_queue<int> que;

    ll ans = 0;
    reps(i,1,M){
        rep(j,0,SZ(v[i])){
            que.push(v[i][j]);
        }
        if(!que.empty()){
            ans += que.top();
            que.pop();
        }
    }
    cout << ans << enld;
    return 0;
}