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>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename Q_temp>
using smaller_queue = priority_queue <Q_temp, vector<Q_temp>, greater<Q_temp> >;

const ll MOD = (ll) (1e9 + 7);
const int INF = (int) 1e9;
const ll LINF = (ll) 4e18;
const double PI = acos(-1.0);

#define REP(i,m,n) for(ll i = m; i < (ll)(n); ++i)
#define rep(i,n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbg2(x_) for(auto a_ : x_) cerr << a_ << " "; cerr << endl;
#define dbg3(x_ , sx_) rep(i, sx_) cerr << x_[i] << " "; cerr << endl;
vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

inline ll CEIL(ll a, ll b) {
    return (a + b - 1) / b;
}

void finish() {
    cout << -1 << endl;
    exit(0);
}

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

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(20) << setiosflags(ios::fixed);

    int n, m;
    cin >> n >> m;
    vector<pll> work;

    rep(i, n) {
        pll p;
        cin >> p.first >> p.second;
        work.push_back(p);
    }

    sort(all(work));

    ll sum = 0;
    priority_queue<ll> que;
    int index = 0;

    rep(i, m) {
        while (index < n && work[index].first <= i + 1) {
            que.push(work[index].second);
            index++;
        }
        if (que.empty()) continue;
        sum += que.top();
        que.pop();
    }
    cout << sum << endl;

    return 0;
}