C. Planning

C. Planning

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n flights that must depart today, the i-th of them is planned to depart at the i-th minute of the day.

Metropolis airport is the main transport hub of Metropolia, so it is difficult to keep the schedule intact. This is exactly the case today: because of technical issues, no flights were able to depart during the first k minutes of the day, so now the new departure schedule must be created.

All n scheduled flights must now depart at different minutes between (k + 1)-th and (k + n)-th, inclusive. However, it's not mandatory for the flights to depart in the same order they were initially scheduled to do so — their order in the new schedule can be different. There is only one restriction: no flight is allowed to depart earlier than it was supposed to depart in the initial schedule.

Helen knows that each minute of delay of the i-th flight costs airport c_i burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 300 000), here n is the number of flights, and k is the number of minutes in the beginning of the day that the flights did not depart.

The second line contains n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 10⁷), here c_i is the cost of delaying the i-th flight for one minute.

Output

The first line must contain the minimum possible total cost of delaying the flights.

The second line must contain n different integers t₁, t₂, ..., t_n (k + 1 ≤ t_i ≤ k + n), here t_i is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.

Example

Input

5 2
4 2 1 10 2

Output

20
3 6 7 4 5 

Note

Let us consider sample test. If Helen just moves all flights 2 minutes later preserving the order, the total cost of delaying the flights would be (3 - 1)·4 + (4 - 2)·2 + (5 - 3)·1 + (6 - 4)·10 + (7 - 5)·2 = 38 burles.

However, the better schedule is shown in the sample answer, its cost is (3 - 1)·4 + (6 - 2)·2 + (7 - 3)·1 + (4 - 4)·10 + (5 - 5)·2 = 20 burles.

 这个可以用优先队列,逐个去除.

#include <bits/stdc++.h>
#define N 300005
#define ll long long int
using namespace std;
struct Node{
  ll pre,cost;
  friend bool operator<(Node a,Node b){
        return a.cost<b.cost;
    }
}node[N];
ll k[N];
ll n,m;
priority_queue<Node> q;
int main(){
    cin>>n>>m;
    for(int i=0;i<n;i++){
      cin>>node[i].cost;
      node[i].pre=i+1;
    }
    int ans=0;
    ll sum=0;
    for(int i=m+1;i<=m+n;i++){
      while(ans<n&&node[ans].pre<=i){
        q.push(node[ans]);
        ans++;
      }
      Node cnt=q.top();
      q.pop();
      k[cnt.pre]=i;
      sum+=(ll)cnt.cost*(i-cnt.pre);
    }
    cout<<sum<<endl;
    for(int i=1;i<=n;i++){
      cout<<k[i]<<" ";
    }
    cout<<endl;
    return 0;
}