Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)C. Planning

time limit per test

1 second

memory limit per test

512 megabytes

Helen works in Metropolis airport. She is responsible for creating a departure schedule. There aren 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 firstk 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), heren 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 integersc₁, c₂, ..., c_n (1 ≤ c_i ≤ 10⁷), here c_i is the cost of delaying thei-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 thei-th flight must depart. If there are several optimal schedules, print any of them.

Example

Input

5 24 2 1 10 2

Output

203 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.

题目思路：贪心法的应用。根据公式 ∑(time-id)cost(k<time<=k+n,time为实际起飞的时间，id为原计划起飞的时间)可知，对于满足条件的(id,cost)组合，应该尽量让cost大的先飞（由于id*cost为定值）。

注意：题目斜体部分的理解，不能提前飞，就是在time时间点只能飞id小的，以及优先队列的使用（第一次写blog）。

题目代码：

#include<bits/stdc++.h> using namespace std;typedef long long ll;struct node{int id,cost;node(int id,int cost):id(id),cost(cost){};bool operator<(const node& n)const{return cost<n.cost;}};int a[300001];int main(){int n,k;priority_queue<node>q;while(~scanf("%d%d",&n,&k)){while(!q.empty()) q.pop();int t;ll sum=0;for(int i=1;i<=n+k;i++){if(i<=n){scanf("%d",&t);q.push(node(i,t));}if(i>k){node tt=q.top();q.pop();sum+=(i-tt.id)*tt.cost;a[tt.id]=i;}}printf("%I64d\n",sum);for(int i=1;i<n;i++) printf("%d ",a[i]);printf("%d\n",a[n]);}return 0;}