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<iostream>
using namespace std;
typedef long long ll;

int main(){
    int n, d;
    cin >> n >> d;
    int t, f, bt = 0, bf = 1;
    ll ans = 0, num = 0;
    while(n-- > 0){
        cin >> t >> f;
        if(t-bt < abs(f-bf)){
            ans = -1;
            break;
        }
        if(bf-1 + f-1 <= t-bt){
            ans += num * (bf-1);
            num = 0;
        }else if(num == d){
            ans = -1;
            break;
        }
        ans += num * abs(t-bt);
        num++;
        bt = t, bf = f;
    }
    if(ans != -1)   ans += num * (bf-1);
    cout << ans << endl;
    return 0;
}