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: Accepted)

#include <bits/stdc++.h>
#ifndef LOCAL
#pragma GCC optimize(3)
#else
#endif
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
using namespace std;
#define ls (rt<<1)
#define rs (rt<<1|1)
typedef long long ll;
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+100;
struct node {
    int a, b;
    bool operator < (const node& aa) const {
        // if (a!=aa.a) return a>aa.a;
        return b>aa.b;
    }
    void in() {
        cin >> a >> b;
    }
}e[maxn];
bool vis[maxn];
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif // LOCAL
    IOS;
    int n, m;
    cin >> n >> m;
    for (int i=0; i<n; i++) {
        e[i].in();
    }
    sort(e, e+n);
    ll ans=0;
    set<int> fr;
    for (int i=0; i<m; i++) fr.insert(i);
    for (int i=0; i<n; i++) {
        if (fr.size()==0) break;
        if (e[i].a<=m) {
            if (!fr.count(m-e[i].a)) {
                set<int>::iterator plc=fr.lower_bound(m-e[i].a);
                if (plc==fr.begin()) continue;
                plc--;
                fr.erase(plc); ans+=e[i].b;
            } else {
                ans+=e[i].b; fr.erase(m-e[i].a);
            }
        }
    }
    cout << ans;
    return 0;
}