Codeforces Round #433 C. Planning（貪心）

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 ci 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, ..., cn (1 ≤ ci ≤ 107), here ci 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, ..., tn (k + 1 ≤ ti ≤ k + n), here ti 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 = 20burles.

題意：原本有n個航班，他們的起飛時間是1-n，現在機場規定在每一天的前k分鐘不能有飛機起飛，那麼就得有航班起飛要延誤，現在給出每個航班延誤一分鐘所消耗的費用，問你怎麼安排飛機的起飛才能使花費最少，飛機起飛時間不能比原本的要早。

用貪心的思想，延誤花費大的肯定要放到前面，而且又不能比原本的時間早，於是我把都延遲k分鐘的時間都放到了set裏，然後每次找不小於原本時間的第一小的時間，這樣就能保證花費最小。

#include<set>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;


struct node
{
    int id,c;
};

node a[300010];

bool cmp(node a,node b)
{
    if(a.c == b.c)
        return a.id < b.id;
    return a.c > b.c;
}

int ans[300010];
set<int> s;
int main(void)
{
    int n,k,i,j;
    while(scanf("%d%d",&n,&k)==2)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i].c);
            a[i].id = i;
            s.insert(k+i);
        }
        sort(a+1,a+1+n,cmp);
        LL sum = 0;
        for(i=1;i<=n;i++)
        {
            int p = *s.upper_bound(a[i].id-1);
            ans[a[i].id] = p;
            sum += (LL)(p-a[i].id)*a[i].c;
            s.erase(p);
        }
        printf("%I64d\n",sum);
        for(i=1;i<=n;i++)
            printf("%d ",ans[i]);
        printf("\n");
    }

    return 0;
}