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)

#define _CRT_SECURE_NO_WARNINGS
#include <math.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <iterator>
#include <vector>
#include <string>
#include <string.h>
#include <queue>
#include <stack>
#include <set>
#include <bitset>
#include <map>
#include <regex>
#include <sstream>
#include <utility>
#include <list>
#include <unordered_set>
#include <unordered_map>
#define endl	'\n'
#define MOD 1000000

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pii;
int dx[] = { 0,1,0,-1 };
int dy[] = { -1,0,1,0 };
int dz[] = { 0,0,0,0,1,-1 };
const int INF = 1000000000;

int n, m;

struct tt {
	int days, reward;
};

bool cmp(const tt &a, const tt &b) {
	if (a.reward != b.reward) return a.reward > b.reward;
	return a.days > b.days;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	
	cin >> n >> m;
	int a, b;
	vector<tt> v;
	for (int i = 0; i < n; i++) {
		cin >> a >> b;
		v.push_back({ a,b });
	}

	sort(v.begin(), v.end(), cmp);
	
	ll ans = 0;
	for (int i = 0; i < m; i++) {
		int remain = m - i;
		int idx = 0;
		for (; idx < v.size() && v[idx].days > remain; idx++);
		if (idx == v.size()) break;
		ans += v[idx].reward;
		v.erase(v.begin() + idx);
		if (v.empty()) break;
	}

	cout << ans;

}