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, n) for (int i = 0; i < (int)n; ++i)
#define REP(i, a, b) for (int i = (int)a; i < (int)b; ++i)
#define each(i,c) for(__typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define all(v) v.begin(), v.end()
#define mset(a, n) memset(a, n, sizeof(a))

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

const int INF = 1000000000;
const int MOD = 1000000007;

bool job_comp(const PAIR &l, const PAIR &r) {
    if (l.second != r.second) return l.second > r.second;
    else return l.first > r.first;
}

int main(int argc, char *argv[]) {
    // cin.tie(0);
    // ios::sync_with_stdio(false);
    int N, M; cin >> N >> M;
    vector<PAIR> job(N);
    rep(i, N) {
        int a, b; cin >> a >> b;
        job[i] = make_pair(a, b);
    }

    sort(all(job), job_comp);

    int ans = 0;
    rep(i, N) {
        if (job[i].first <= M) ans += job[i].second;
        else break;
        M--;
    }
    cout << ans << endl;
    return 0;
}