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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i <= n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VSORTR(v) sort(v.rbegin(), v.rend());
#define ALL(v) (v).begin(),(v).end()
 
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;

template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

const ll MOD = 998244353;
const ll inf = 1e18;

int main(){
	cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    ll n, m;
    cin >> n >> m;
    vector<P> p(n);
    ll a, b;
    REP(i, n) {
        cin >> a >> b;
        p[i] = {b, a};
    }
    VSORTR(p);
    ll res = 0;
    vll used(m + 1);
    REP(i, n) {
        a = p[i].second, b = p[i].first;
        while (used[a] && a <= m) a++;
        if (a > m) continue;
        res += b;
        used[a] = 1;
    }
    cout << res << endl;
    return 0;
}