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;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
typedef long long int ll;

#define EPS (1e-7)
#define INF (1e9)
#define LLINF (1e18)
#define PI (acos(-1))
#define MOD (1000000007)
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n, m;
    cin >> n >> m;
    priority_queue<pair<ll, ll>> pq;
    for (ll i = 0; i < n; i++)
    {
        ll a, b;
        cin >> a >> b;
        pq.push(make_pair(b, a));
    }
    ll ans = 0;

    while (!pq.empty() && m >= 0)
    {
        auto p = pq.top();
        pq.pop();
        if (p.second <= m)
        {
            ans += p.first;
            // cout << "dbg:" << p.first << ' ' << p.second << ' ' << m << endl;
            m--;
        }
        // cout << p.first << ' ' << p.second << endl;
    }
    cout << ans << endl;
}