Codeforces853A Planing （优先队列+贪心）

A. 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

Copy

5 2
4 2 1 10 2

Output

Copy

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.

题解：在k+1到k+n时间内把将要耽误的飞机加入优先队列中，自大到小排序，优先让花费大先起飞
注意 数据范围 输出long long

//优先队列
//默认从大到小
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<set>
#include<queue>
#define ll long long
using namespace std;
const int maxn = 300000+5;
struct node
{
    int i,x;
    bool operator < (const node &A) const
    {
        return A.x>x;
    }
}a[maxn];
int n,k;

int main()
{

    scanf("%d %d",&n,&k);
    int c;
    priority_queue<node>s;
    for(int i=1;i<=n;i++){
        scanf("%d",&c);
        a[i].x=c;
        a[i].i=i;
    }
    ll sum=0;
    int t=k;//标记误机时间
    //for(int i=1;i<=n;i++){
     //   s.push(a[i]);
    //}
    t=k+1;
    int ans=1;
    for(int i=1;i<=n;i++){
        if(i<=t){
            //会耽误飞机
            s.push(a[i]);
        }
        else if(i>t){
            sum+=(long long)(t-s.top().i)*s.top().x;
           // printf("t-%d-x-%d-sum-%d\n",t,s.top().x,sum);
            a[s.top().i].i=t;
            s.pop();
            t++;
            --i;
        }
    }
    while(!s.empty()){
        sum+=(long long)(t-s.top().i)*s.top().x;
        a[s.top().i].i=t;
        s.pop();
        t++;
    }
    printf("%lld\n",sum);
    for(int i=1;i<=n;i++){
        printf("%d ",a[i].i);
    }
    return 0;
}