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;

int main(){
  
  int N,M;
  cin >> N >> M;
  vector<int> A(N),B(N);
  
  for (int i=0;i<N;i++){
    int a,b;
    cin >> a >> b;
    A[i] = a;
    B[i] = b;
  }
  
  vector<vector<int>> dp(N+1, vector<int>(N+10));
  for (int d=0;d<N+10;d++){
    dp[0][d] = 0;
  }
  
  for (int i=0;i<N;i++){
    for (int d=0;d<N+10;d++) {
      if (d >= 1) dp[i+1][d] = max(dp[i][d-A[i]] + B[i], dp[i][d]);
      else dp[i+1][d] = dp[i][d];
    }
  }
  
  for (int i=0;i<N+1;i++){
    for (int d=0;d<N+10;d++) {
    }
  }
  
  cout << dp[N][M+1] << endl;
}