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

C. Planning

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

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 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 = 20burles.

题目大意:现在有n架飞机本来按顺序起飞，现在飞机时间都在k分钟后起飞，且飞机的先后顺序可以改变，对于每个飞机来说，若是他晚点飞会有个花费，现在要求找出一个起飞顺序，花费最小

解题思路:我现在能做的是，按照花费将飞机排序，对于当前未安排起飞的飞机来说，若是能够不晚点飞是最好的，但是若不能准点起飞，那么我就要找到最早起飞的顺序给他，这个时间复杂度就上来了，能够做的就是用并查集，以类似链子的样子连起来

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>using namespace std;typedef long long LL;struct point{int c,index;}f[300005];int a[300005];int check[800005];int n,k,i,cnt;LL sum;bool cmp(point x,point y){return x.c>y.c;}int find(int x){if(check[x]==x)return x;check[x]=find(check[x]);return check[x];}int main(){cin>>n>>k;for(i=1;i<=n;i++){cin>>f[i].c;f[i].index=i;}sort(f+1,f+1+n,cmp);for(i=k+1;i<=k+n;i++) check[i]=i;sum = 0;for(i=1;i<=n;i++){cnt = max(f[i].index,k+1);a[f[i].index]=find(cnt);sum += (LL)(a[f[i].index]-f[i].index)*f[i].c;check[a[f[i].index]]=find(a[f[i].index]+1);}cout<<sum<<endl;for(i=1;i<n;i++) cout<<a[i]<<" ";cout<<a[i]<<endl;return 0;}