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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'
#define mset(d,val) memset(d,val,sizeof(d))
#define setp(x) cout<<fixed<<setprecision(x)
#define forn(i,a,b) for(int i=(a);i<(b);i++)
#define fore(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define F first
#define S second
#define pqueue priority_queue
#define fbo find_by_order
#define ook order_of_key
typedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef long double ld;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;
void amin(ll &a, ll b){ a=min(a,b); }
void amax(ll &a, ll b){ a=max(a,b); }
void YES(){cout<<"YES\n";} void NO(){cout<<"NO\n";}
void SD(int t=0){ cout<<"PASSED "<<t<<endl; }
const ll INF = ll(1e18);
const int MOD = 998244353;

bool DEBUG=0;
const int MAXN = 100005;

ll a[MAXN],b[MAXN];
vi eve[MAXN];
multiset<ll,greater<ll>> s;

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	int n,m; cin>>n>>m;
	forn(i,0,n){
		cin>>a[i]>>b[i];
		if(a[i]>m) continue;
		eve[m-a[i]].pb(b[i]);
	}
	ll ans=0;
	
	for(int i=m-1;i>=0;i--){
		for(ll x: eve[i]){
			s.insert(x);
		}
		if(s.size()){
			ans+=*s.begin();
			s.erase(s.begin());
		}
	}
	cout<<ans<<'\n';
	
	return 0;
}