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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MOD = 1000000007;

struct UniTree {
    int *par, *rank;

    UniTree(int n) {
        par = new int[n];
        rank = new int[n];
        for(int i = 0; i < n; i++) {
            par[i] = i;
            rank[i] = 0;
        }
    }

    int root(int x) {
        if(par[x] == x) return x;
        else return par[x] = root(par[x]);
    }

    void merge(int x, int y) {
        x = root(x);
        y = root(y);
        if(x == y) return;
        if(rank[x] < rank[y]) {
            par[x] = y;
        }else {
            par[y] = x;
            if(rank[x] == rank[y]) rank[x]++;
        }
    }
    
    bool same(int x, int y) {
        return root(x) == root(y);
    }
};

int main() {
    int n, m;
    cin >> n >> m;
    priority_queue<pair<int, int> > pq;
    rep(i, n) {
        int a, b;
        cin >> a >> b;
        pq.push(make_pair(b, a));
    }

    ll ans = 0;
    int day = 0;
    while(!pq.empty()) {
        P p = pq.top();
        pq.pop();
        if(day + p.second <= m) {
            ans += p.first;
            day++;
        }
    }

    cout << ans << endl;

    return 0;
}