CodeForces - 854C Planning（貪心 + 優先隊列）

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 = 20 burles.

 

題意：

共n個航班，預計出發時間爲1， 2， 3， ……（分鐘）， 現在所有航班出發時間延遲k分鐘，第i個航班延遲1分鐘損失ci元，現在可以調整航班出發順序，使得總損失最小。（每架航班不能早於預定時間）

思路：

枚舉時間（1~n），在該時間可以出發的所有航班中選擇開銷最大的航班。

用優先隊列記錄某個時間點可以出發的航班，按開銷從大到小排序

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e5 + 20;

struct node
{
    int num;
    ll cost;
    bool operator < (const node &a)const
    {
        return cost < a.cost;
    }
}s[N];

int main()
{
    int n, k;
    while(~scanf("%d%d", &n, &k))
    {
        priority_queue<node>q;
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &s[i].cost);
            s[i].num = i;
        }
        for(int i = 1; i <= k; ++i)  ///初始時間 <= k的肯定可以起飛
        {
            q.push(s[i]);
        }
        ll ans = 0;
        for(int i = k + 1; i <= k + n; ++i)     
        {
            if(i <= n)
            {
                q.push(s[i]);
            }
            node tmp = q.top();
            q.pop();
            ans += tmp.cost *(i - tmp.num);
            s[tmp.num].num = i;     ///修改起飛時間
        }
        cout<<ans<<'\n';
        for(int i = 1; i <= n; ++i)
        {
            cout<<s[i].num;
            if(i < n)
                cout<<' ';
        }
        cout<<'\n';
    }
    return 0;
}