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 <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <vector>

const int MOD = 1e9 + 7;
const int iINF = 2147483647 / 2;
const long long int llINF = 9223372036854775807 / 2;

using namespace std;
using ll = long long int;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;

typedef pair<ll, ll> pll;
bool paircomp(const pll &a, const pll &b) {
  if (a.second == b.second)
    return a.first > b.first;
  return a.second > b.second;
}

#define REP(i, n) for (ll i = 0; i < (n); i++)
#define RREP(i, n) for (ll i = (n)-1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define AUTO(i, m) for (auto &i : m)
#define ALL(a) (a).begin(), (a).end()
#define MAX(vec) *std::max_element(vec.begin(), vec.end())
#define MIN(vec) *std::min_element(vec.begin(), vec.end())
#define ARGMAX(vec)                                                            \
  std::distance(vec.begin(), std::max_element(vec.begin(), vec.end()))
#define ARGMIN(vec)                                                            \
  std::distance(vec.begin(), std::min_element(vec.begin(), vec.end()))
#define REV(T) greater<T>()
#define PQ(T) priority_queue<T, vector<T>, greater<T>>
#define VVL(a, b, c) vector<vector<ll>>(a, vector<ll>(b, c))
#define VVVL(a, b, c, d)                                                       \
  vector<vector<vector<ll>>>(a, vector<vector<ll>>(b, vector<ll>(c, d)))
#define SP(a) setprecision(a)
#define SQRT(a) sqrt((long double)(a))
#define DPOW(a, b) pow((long double)(a), (long double)(b))
#define UNIQUE(vec)                                                            \
  do {                                                                         \
    sort(ALL((vec)));                                                          \
    (vec).erase(std::unique(ALL((vec))), (vec).end());                         \
  } while (0)

ll POW(ll n, ll m) {
  if (m == 0) {
    return 1;
  } else if (m % 2 == 0) {
    ll tmp = POW(n, m / 2);
    return (tmp * tmp);
  } else {
    return (n * POW(n, m - 1));
  }
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll N, M;
  cin >> N >> M;
  vector<pll> C(N);

  REP(i, N) {
    ll a, b;
    cin >> a >> b;
    C[i] = make_pair(b, a);
  }
  sort(ALL(C), REV(pll));

  // priority_queue<pll> D;
  // REP(i, min(N, M)) { D.push(make_pair(C[i].second, C[i].first)); }
  // ll now = D.size();
  ll ans = 0;
  vl D(M, -1);
  vl DP(N, -1);

  AUTO(c, C) {
    ll now = M - c.second;
    if (DP[now] != -1)
      now = DP[now];
    while (now >= 0) {
      if (D[now] == -1) {
        D[now] = 1;
        ans += c.first;
        DP[M - c.second] = now;
        break;
      } else {
        now--;
      }
    }
  }

  /*REP(i, now) {
    pll po = D.top();
    D.pop();
    cout << po.first << "," << po.second << endl;
  }*/
  cout << ans << endl;
  return 0;
}