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)

#include <bits/stdc++.h>
//#pragma GCC optimize(2)

using namespace std;

#define LL long long
#define Re register
#define U unsigned
#define forn(i,a,b) for(int i = a;i <= b;++i)
#define nfor(i,a,b) for(int i = a;i >= b;--i)
#define CLR(i,a) memset(i,a,sizeof(i))
#define BR printf("--------------------\n")
#define pb push_back
#define Rep(i,a,b) for(int i=a;i<=b;i++)
#define DEBUG(x) std::cerr << #x << '=' << x << std::endl
const double PI=atan(1.)*4.;
const int maxn=1e5+5;
const int inf = 0x3f3f3f3f;
const int mod=1e9+7;
const double e=2.71828182845904523536;
int dirx[8]={1,0,-1,0,1,-1,-1,1};
int diry[8]={0,1,0,-1,1,1,-1,-1};
int n,m;
struct node
{
    LL a,b;
    double x;
}st[maxn];

bool cmp(node a,node b)
{
    return a.a^b.a?a.a<b.a:a.b>b.b;
}


int main()
{
	ios::sync_with_stdio(false);
    cin.tie(0);
   	cout.tie(0);
    cin>>n>>m;
    for(Re int i=0;i<n;i++)
    {
        cin>>st[i].a>>st[i].b;
    }
    sort(st,st+n,cmp);
    LL ans=0;
    priority_queue<int> q;
    int cur=0;
    for(Re int i=m-1;i>=0;i--)
    {
        while(cur<n&&i+st[cur].a<=m)
        {
            q.push(st[cur++].b);
        }
        if(!q.empty())
        {
            ans+=q.top();
            q.pop();
        }
    }
    cout<<ans;
	return 0;
}