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;
typedef long long ll;
#define int ll
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a) (a).begin(),(a).end()
#define pb emplace_back

signed main(){
    int n,d;
    cin>>n>>d;
    vector<int> f(n), t(n);
    rep(i,n){
        cin>>t[i]>>f[i];
        f[i]--;
    }
    
    int nowt = 0, nowf = 0, num = 0;
    
    int sum = 0;
    rep(i,n){
        if(abs(f[i]-nowf)>t[i]-nowt){
            cout<<-1<<endl;
            return 0;
        }
        if(f[i]+nowf<=t[i]-nowt){
            sum += num*nowf;
            num = 1;
            nowt = t[i];
            nowf = f[i];
        }else{
            if(num+1>d){
                cout<<-1<<endl;
                return 0;
            }
            sum += (t[i]-nowt)*num;
            num ++;
            nowt = t[i];
            nowf = f[i];
        }
    }
    sum += num*nowf;
    cout<<sum<<endl;
}