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>
#define ms(x, n) memset(x,n,sizeof(x));
#define rep(i, n) for(int i = 0; i < (int)n; ++i)
#define rep1(i, n) for(int i = 1; i <= (int)n; ++i)
#define pb push_back
#define PI acos(-1.0)
#define LONG_LONG_MAX 9223372036854775807
#define LONG_LONG_MIN -9223372036854775808
typedef long long int ll;
typedef unsigned long long int ull;
using namespace std;

const int maxn = 1e5 + 10;
int n, m, a, b;
vector<int> vec[maxn];
multiset<int> mts;

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m;
    rep(i, n) {
        cin >> a >> b;
        if (m - a >= 0) vec[m-a].pb(b);
    }
    ll ans = 0;
    for (int i = m; i >= 0; --i) {
        while (!vec[i].empty()) {
            mts.insert(*prev(vec[i].end()));
            vec[i].pop_back();
        }
        if (!mts.empty()) {ans += *prev(mts.end());
        mts.erase(prev(mts.end()));}
    }
    cout << ans << endl;
    return 0;
}