【贪心】Codeforces Round #433

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.

Examples

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.

Solution

题意：原本有n个航班分别将于第1...n分钟起飞，现在由于技术问题第1...k分钟不能有任何航班起飞，而第i个航班每延误1分钟机场就会赔偿ci元，现在问你如何安排新的航班起飞时刻能使得总的赔偿金额最小。每分钟只能有一架飞机起飞且新的起飞顺序不必跟原来的相同，但新的起飞时刻不能早于原来的起飞时刻。

思路：可以证明先将ci大的航班排在尽可能前面会使得总赔偿最少（考虑反证法，证明略）。所以实现时就只需要每次求个mex（minimum excluded，最小不在集合中的非负整数）并判断是否大于等于原来的起飞时刻就好了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 struct Node{
 5     int c,id;
 6     bool operator < (const Node & x) const{
 7         if(c != x.c) return c > x.c;
 8         return id < x.id;
 9     }
10 }p[300033];
11 
12 int n,k,c,mex,t[300033];
13 bool vis[10001000];
14 
15 int getMex(){
16     for(int i = mex;i <= 10000000;++i) if(!vis[i]){
17         mex = i;
18         return i;
19     }
20 }
21 
22 int main(){
23     cin >> n >> k;
24     for(int i = 1;i <= n;++i){
25         scanf("%d",&c);
26         p[i].c = c, p[i].id = i;
27     }
28     sort(p+1,p+1+n);
29     for(int i = 0;i <= k;++i) vis[i] = 1;
30     mex = k + 1;
31     long long ans = 0;
32     for(int i = 1;i <= n;++i){
33         int m = getMex();
34         if(m < p[i].id){
35             vis[p[i].id] = 1;
36             t[p[i].id] = p[i].id;
37         }
38         else{
39             vis[m] = 1;
40             ans += 1LL * p[i].c * (m - p[i].id);
41             t[p[i].id] = m;
42         }
43     }
44     cout << ans << endl;
45     for(int i = 1;i <= n;++i) printf("%d ",t[i]);
46 
47     return 0;
48 }