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 <algorithm>
#include <array>
#include <bitset>
#include <deque>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>
//region cp_helper
using namespace std;
#define _io ios_base::sync_with_stdio(false); cin.tie(nullptr);
#define ll long long
#define ull unsigned long long
#define ii pair<int, int>
#define vi vector<int>
#define vvi vector< vector<int> >
#define vll vector<ll>
#define vii vector<ii>
#define in(n) cin >> n
#define out(n) cout << n
#define outs(n) cout << n << ' '
#define outn(n) cout << n << '\n'
#define vin(a, n) for (int i=0; i<n; i++) in(a[i])
#define viin(a, n) for (int i=0; i<n; i++) in(a[i].first) >> a[i].second;
#define vout(a) for (auto i : a) outs(i)
#define all(a) a.begin(), a.end()
#define qsort(a) sort(all(a))
#define qdsort(a) sort(all(a), greater<>())
const int mod = (int)(1e9) + 7;
//endregion

using namespace std;
int n, m, s[100005], c, total;
int main() { _io
    in(n); in(m);
    vii arr(n);
    viin(arr, n);
    sort(all(arr), [](ii a, ii b) {
        if (a.second == b.second) return a.first < b.first;
        return a.second>b.second;
    });
    for (auto i : arr) {
        if ((m-i.first) >= s[i.first]) {
            total += i.second;
            s[i.first]++;
            c++;
        }
        if (c >= m) break;
    }
    outn(total);
}