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;
#define REP(i,a) for(int i = 0; i < (a);i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const int MOD = 1e9 + 7;
using Graph = vector < vector<int>>;
/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x,long long int y) {
	return(x + y-1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}

int main() {
	int N, M;
	cin >> N >>M;
	vector<pair<int, int>> s;
	for (int i = 0; i < N; i++) {
		int a;
		int b;
		cin >> a >> b;
		pair<int, int> key;
		key.first = b;
		key.second = a;
		s.push_back(key);
	}
	long long int sum = 0;
	sort(s.begin(), s.end(),greater<pair<int,int>>());
	pair<int, int> key;
	int c = 0;
	int day = 0;
	for (int j = 0; j < N; j++) {
			pair<int, int>key;
			key = s[j];
			int a, b;
			b = key.first;
			a = key.second;
			if (a+day <= M) {
				sum = sum + b;
				c = j+1;
				day = day + 1;
			}
			if (day == M) {
				break;
			}
			if (j == N - 1) {
				break;
			}
	}
	cout << sum;
}