Codeforces Round #433 C. Planning（贪心）

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.

题意：原本有n个航班，他们的起飞时间是1-n，现在机场规定在每一天的前k分钟不能有飞机起飞，那么就得有航班起飞要延误，现在给出每个航班延误一分钟所消耗的费用，问你怎么安排飞机的起飞才能使花费最少，飞机起飞时间不能比原本的要早。

用贪心的思想，延误花费大的肯定要放到前面，而且又不能比原本的时间早，于是我把都延迟k分钟的时间都放到了set里，然后每次找不小于原本时间的第一小的时间，这样就能保证花费最小。

#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;
}