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)

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/
#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;

inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }


/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
*/
///**
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
  return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if ( it != v.begin() ) os << ", ";
    os << *it;
  }
  return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if ( it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if ( it != v.begin() ) os << ", ";
    os << it -> first << " = " << it -> second ;
  }
  return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () { cerr << endl; }

template <typename T>
void faltu( T a[], int n ) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }
// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
#include <iostream>
using namespace __gnu_pbds;
using namespace std;
// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;

// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।

//*//**___________________________________________________**/

const int N = 1e5 + 5;
vector<int> idx[N];

int main()
{
  FASTIO
  /*
#ifndef ONLINE_JUDGE
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
  freopen("error.txt", "w", stderr);
#endif
//*/
  int n, m;
  cin >> n >> m;
  priority_queue<int> PQ;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    if (m - a >= 0) idx[m - a].push_back(b);
  }
  int ans = 0;
  for (int i = m; i >= 0; i--) {
    for (auto x : idx[i])PQ.push(x);
    if ((int)PQ.size()) {
      ans += PQ.top();
      PQ.pop();
    }
  }
  cout << ans << endl;
  return 0;
}