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)

// ABC137D3.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//
#include <iostream>
#include<vector>
#include <numeric>
#include<algorithm>
using namespace std;

//日数、報酬
struct work {
	int day;
	int pay;
};
int main(int argc, const char* argv[]) {
	int N, M;
	cin >> N >> M;
	vector<work> works(N);
	int A, B;
	for (int i = 0; i < N; i++) {
		cin >> A >> B;
		work w;
		w.day = A;
		w.pay = B;
		works.at(i) = w;
	}
	//報酬の降順で日数の降順
	sort(works.begin(), works.end(), [](const work& x, const work& y) {return x.day > y.day; });
	sort(works.begin(), works.end(), [](const work& x, const work& y) {return x.pay > y.pay; });
	/*
	cout << endl;	
	for (int i = 0; i < N; i++)
	{
		cout << works.at(i).day << " " << works.at(i).pay << endl;
	}
	*/
	int ans = 0;
	int rest = M; //アルバイトできる残りの日数
	for (int j = 0; j < N; j++)
	{
			if (works.at(j).day <= rest)
			{
				ans += works.at(j).pay;
				//cout << j << endl;
				rest--;// 該当するアルバイトがあったら、それを行い、残りの日数を減らす
				if (rest<1)
				{
					break;
				}
			}
	}
	cout << ans << endl;
	return 0;
}