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>
using namespace std;
 
typedef long long ll;
typedef double D;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
 
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d %d",&x,&y)
#define rl(x) scanf("%lld",&x)
#define rl2(x,y) scanf("%lld %lld",&x,&y)
#define wd(x) printf("%d" ,x)
#define wd2(x,y) printf("%d %d",x,y)
#define wl(x) printf("%lld",x)
#define wl2(x,y) printf("%lld %lld",x,y)
#define PC(x) putchar(x)
#define GC() getchar()
#define NL printf("\n")
#define SP printf(" ")
 
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define ITER(it,a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a,value) memset(a, value, sizeof(a))

#define F first
#define S second
#define MP make_pair
#define PB push_back

#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define fr(i,s,e) for(int i=s;i<e;i++)
#define frl(i,s,e) for(ll i=s;i<e;i++)
#define frr(i,s,e) for(int i=s-1;i>=e;i--)
#define frv(i,a) for(int i = 0;i<(int)a.size();i++)
#define frvr(i,a) for(int i = a.size()-1;i>=0;i--)
#define tr(i,a) for(typeof(a.begin()) i = a.begin(); i != a.end();i++)

#define mod 1000000007
#define maxn 100009
#define maxr 100009
int main(){
	//sstd::ios::sync_with_stdio(false); cin.tie(NULL);
	ll n,m;rl2(n,m);
	vector<PLL> arr;
	priority_queue<PLL> q;

	fr(i,0,n)
	{
		ll x,y;
		rl2(x,y);
		arr.PB(MP(x,y));
		q.push(MP(y,x));
	}
	sort(ALL(arr));
	ll ans=0;
	map<ll,ll> mp;
	int flag=1;
	// fr(i,0,n)
	// {
	// 	if(arr[i].F>m)
	// 		break;
	// 	if(!mp.count(arr[i].F))
	// 	{
	// 		ans+=arr[i].S;
	// 		mp[arr[i].F]=arr[i].S;
	// 	}
	// 	else
	// 	{
	// 		if(arr[i].S>mp[arr[i].F])
	// 		{
	// 			ans-=mp[arr[i].F];
	// 			mp[arr[i].F]=arr[i].S;
	// 			ans+=arr[i].S;
	// 		}
	// 	}
	// }
	int i=0;
	while(i++<m)
	{
		PLL pr=q.top();
		q.pop();
		if(pr.S<=m)
		{
			// if(pr.S==m)
			// {
			// 	if(flag)
			// 		ans+=pr.F;
			// }
			// else
			// 	ans+=pr.F;
			// if(pr.S==m)
			// {
			// 	flag=0;
			// }
			if(mp[pr.S]<m-pr.S+1)
			{
				mp[pr.S]++;
				ans+=pr.F;
			}
		}
	}
	wl(ans);
	NL;
	return 0;
}