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 JavaScript (node.js v5.12) (judge verdict: Runtime Error)



function Main(input) {
  input = input.split("\n");
  var N = parseInt(input[0].split(" ")[0]);
  var M = parseInt(input[0].split(" ")[1]);
  input.shift();
  input.splice(N,input.length);
  var x = input.map(i => i.split(" ").map(i=> parseInt(i)));
  var dp=new Array(Math.pow(10,5)*N+1);
  for(var i=0;i<dp.length;i++){
    if(i==0){
      var sub = new Array(M+1).fill(1);
    }else{
      var sub = new Array(M+1).fill(0);
    }
    dp[i]=sub;
  }
  x.sort(function(a,b){
    if (a[0] < b[0]) return -1;
    if (a[0] > b[0]) return 1;
    if (a[1] < b[1]) return -1;
    if (a[1] > b[1]) return 1;
    return 0;
  })
  dp[0][0]=1;
  for(k=0;k<N;k++){
    for(i=M;i>=1;i--){
      for(j=dp.length-1;j>=0;j--){
        if(dp[j][i-1]==1){
          dp[j][i]=1;
          if(i>=x[k][0]){
            dp[j+x[k][1]][i]=1;
          }
        }
      }
    }
  }
  for(var i=dp.length-1;i>=0;i--){
    if(dp[i][M]==1){
      var ans = i;
      break;
    }
  }

  console.log('%s', ans);
  return ans;
}

function debug(n){
  var input = document.getElementById("input"+n).value;
  var ans=Main(input);
  var result="WA";
  if(ans==document.getElementById("output"+n).value.split("\n")[0]){
    result="AC";
  }
  document.getElementById("result"+n).innerHTML=result;
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));