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>
using namespace std;
long long a[100010] , b[30] , INF=100000000007 , ans=0 , x=0;
/*long long f(long long k)
{
	long long c=((k+99)*(k+10))%INF;
	return c*(c+21)%INF;
}*/
int main()
{
	for(int i=1;i<=28;i++)
	{
		b[i]=181384*i*i-i;
		b[i]%=INF;
		b[i]*=(b[i]+i);
		b[i]%=INF;
		b[i]-=(b[i]/i)-INF;
		b[i]+=i*i*i;
		b[i]%=INF;
	}
	int n;	cin>>n;
	for(int i=1;i<=n;i++)
	{
		string s;	cin>>s;
		for(int j=0;j<s.size();j++){
			a[i]+=b[s[j]-'a'+1];
			a[i]%=INF;
		}
	}
	sort(a+1 , a+1+n);
	a[n+1]=-1;
	for(int i=1;i<=n;i++)
	{
		x++;
		if(a[i]!=a[i+1]){
			ans+=x*(x-1)/2;
			x=0;
		}
	}
	/*for(int i=1;i<=n;i++)
		cout<<a[i]<<endl;*/
	cout<<ans<<endl;
}