Problem: problem

AOR Co., Ltd. (Association Of Return home) has $ N $ employees.

Employee $ i $ wants to use the elevator to go down to the $ 1 $ floor and arrives in front of the elevator on the $ F_i $ floor at time $ t_i $. You decide to remotely control an elevator that has only $ 1 $ on the $ 1 $ floor at time $ 0 $ and send all employees to the $ 1 $ floor. The elevator can only carry up to $ D $ people. Elevators can move upstairs or stay in place per unit time. Employee $ i $ only takes the elevator if the elevator is on the $ F_i $ floor at time $ t_i $ and does not exceed capacity. When I can't get on the elevator at time $ t_i $, I'm on the stairs to the $ 1 $ floor. The time it takes to get on and off can be ignored.

Find the minimum total time each employee is in the elevator. However, if there are even $ 1 $ people who cannot take the elevator to the $ 1 $ floor, output $ -1 $.



output

Output the minimum total time each employee is in the elevator. However, if there are even $ 1 $ people who cannot take the elevator to the $ 1 $ floor, output $ -1 $. Also, output a line break at the end.

Example

Input

2 2
2 2
3 3


Output

5

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int main(){
  long long n,d,t[100000],f[100000],nf=1,nt=0,nd=0,ans=0;
  cin>>n>>d;
  for(int i=0;i<n;i++){
    scanf("%lld %lld",&t[i],&f[i]);
  }
  for(int i=0;i<n;i++){
    long long dist=abs(f[i]-nf);
    long long T=t[i]-nt;
    if(T<dist){
      cout<<-1<<endl;
      return 0;
    }
    if(nf-1+f[i]-1<=T){
      ans+=nd*(nf-1);
      nd=0;
    }else{
      ans+=nd*T;
    }
    nd++;
    if(nd>d){
      cout<<-1<<endl;
      return 0;
    }
    nf=f[i];
    nt=t[i];
  }
  ans+=nd*(nf-1);
  cout<<ans<<endl;
  return 0;
}