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)

//#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std; typedef long double ld; typedef long long ll;
typedef unsigned long long ull;
#define endl "\n"
#define MP make_pair
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define FORR(x,arr) for(auto& x:arr)
#define PII pair<int, int>
#define FI first 
#define SE second
#define ALL(x) (x).begin(), (x).end()
const int INF=1<<30; const ll LINF=1LL<<60; const ll mod=1e9+7; const int NIL = -1;
template<class T>inline bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>inline bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//-------------------
vector<pair<int, int>> v;

int main(){
	int n,m; cin >> n >> m;
	priority_queue<int> que;
	FOR(i,1,n){
		int a,b; cin >> a >> b;
		v.push_back(MP(a,b));
	}
	sort(ALL(v));
	int cur = 0;
	ll ans = 0;

	FOR(i,1,m){
		while(v[cur].first<=i && cur <= n-1){
			que.push(v[cur].second);
			cur++;
		}
		if(!que.empty()){
			ans += que.top(); que.pop();
		}
	}
	cout << ans << endl;
    return 0;
}