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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<utility>
#include<set>
#include<stack>
#include<list>
#include<deque>
#include<bitset>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<climits>
#include<cmath>
#include<cctype>


#define pb push_back
#define mp make_pair
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ren(i,a,b) for(int i=a;i>=b;i--)
#define ff first
#define ss second
#define pll pair<long long int,long long int>
#define pii pair<int,int>
#define vll vector<long long int>
#define vii vector<int>
#define gi(n) scanf("%d",&n)
#define gll(n) scanf("%lld",&n)
#define gstr(n) scanf("%s",n)
#define gl(n) cin >> n
#define oi(n) printf("%d",n)
#define oll(n) printf("%lld",n)
#define ostr(n) printf("%s",n)
#define ol(n) cout << n
#define os cout<<" "
#define on cout<<"\n"
#define o2(a,b) cout<<a<<" "<<b
#define all(n) n.begin(),n.end()
#define present(s,x) (s.find(x) != s.end())
#define cpresent(s,x) (find(all(s),x) != s.end())
#define tr(container, it) for(__typeof(container.begin()) it = container.begin(); it != container.end(); it++)
using namespace std;

typedef unsigned long long int ull;
typedef long long int ll;
typedef long double ld;
typedef vector<vector<ll> > mat;

pll p[200005];

int main()
{ios_base::sync_with_stdio(false);
int n,m;
cin>>n>>m;

rep(i,0,n-1)cin>>p[i].ss>>p[i].ff;

ll ans=0;

rep(i,0,n-1)p[i].ss=-p[i].ss,p[i].ff=-p[i].ff;

sort(p,p+n);

ll taken=0;

rep(i,0,n-1)
{
	ll cost=-p[i].ff,days=-p[i].ss;
	//o2(cost,taken+days);on;
	if(taken+days<=m)
	{
		taken++;
		ans+=cost;
	}
}
ol(ans);

return 0;
}