B. Yet Another Array Partitioning Task ——cf

B. Yet Another Array Partitioning Task

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

An array bb is called to be a subarray of aa if it forms a continuous subsequence of aa , that is, if it is equal to alal , al+1al+1 , …… , arar for some l,rl,r .

Suppose mm is some known constant. For any array, having mm or more elements, let's define it's beauty as the sum of mm largest elements of that array. For example:

• For array x=[4,3,1,5,2]x=[4,3,1,5,2] and m=3m=3 , the 33 largest elements of xx are 55 , 44 and 33 , so the beauty of xx is 5+4+3=125+4+3=12 .
• For array x=[10,10,10]x=[10,10,10] and m=2m=2 , the beauty of xx is 10+10=2010+10=20 .

You are given an array a1,a2,…,ana1,a2,…,an , the value of the said constant mm and an integer kk . Your need to split the array aa into exactly kk subarrays such that:

• Each element from aa belongs to exactly one subarray.
• Each subarray has at least mm elements.
• The sum of all beauties of kk subarrays is maximum possible.

Input

The first line contains three integers nn , mm and kk (2≤n≤2⋅1052≤n≤2⋅105 , 1≤m1≤m , 2≤k2≤k , m⋅k≤nm⋅k≤n ) — the number of elements in aa , the constant mm in the definition of beauty and the number of subarrays to split to.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109 ).

Output

In the first line, print the maximum possible sum of the beauties of the subarrays in the optimal partition.

In the second line, print k−1k−1 integers p1,p2,…,pk−1p1,p2,…,pk−1 (1≤p1<p2<…<pk−1<n1≤p1<p2<…<pk−1<n ) representing the partition of the array, in which:

• All elements with indices from 11 to p1p1 belong to the first subarray.
• All elements with indices from p1+1p1+1 to p2p2 belong to the second subarray.
• …… .
• All elements with indices from pk−1+1pk−1+1 to nn belong to the last, kk -th subarray.

If there are several optimal partitions, print any of them.

Examples

Input

Copy

9 2 3
5 2 5 2 4 1 1 3 2

Output

Copy

21
3 5 

Input

Copy

6 1 4
4 1 3 2 2 3

Output

Copy

12
1 3 5 

Input

Copy

2 1 2
-1000000000 1000000000

Output

Copy

0
1 

Note

In the first example, one of the optimal partitions is [5,2,5][5,2,5] , [2,4][2,4] , [1,1,3,2][1,1,3,2] .

• The beauty of the subarray [5,2,5][5,2,5] is 5+5=105+5=10 .
• The beauty of the subarray [2,4][2,4] is 2+4=62+4=6 .
• The beauty of the subarray [1,1,3,2][1,1,3,2] is 3+2=53+2=5 .

The sum of their beauties is 10+6+5=2110+6+5=21 .

In the second example, one optimal partition is [4][4] , [1,3][1,3] , [2,2][2,2] , [3][3] .

大意：
这个题目是给你n个数据，让你分成k段，每一段至少m个数，并且把每一段的前m大的数求和
输出求和的结果，和分段的位置、
思路：
是求这每一段的前m大的数，其实可以转化成，求这一组数前m*k大的数之和。
至于求分段的位置，这个就有点麻烦了，可以把每一个位置初始标记为0，
然后在求sum的同时，将求过的数位置标记成1，之后求分段位置的时候，
就可以进行累加，一旦累加之和为m，说明这m个数组成了一段，也就是在这个位置将数组分段。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
struct node
{
    int x,id;
}a[maxn];
bool vis[maxn];
bool cmp(node a,node b)
{
    return a.x>b.x;
}
int main()
{
    int n,m,k;
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].x);
        a[i].id=i;
    }
    sort(a+1,a+1+n,cmp);
    ll sum=0;
    for(int i=1;i<=m*k;i++)
    {
        sum+=a[i].x;
        vis[a[i].id]=1;
        //printf("a.id=%d\n",a[i])
    }
    printf("%I64d\n",sum);
    int s=0;;
    for(int i=1;i<=n;i++)
    {
        s=s+vis[i];
      //  printf("%d %d\n",i,s);
        if(s==m&&k!=1)
        {
        	k--;
        	s=0;
            printf("%d ",i);
        }
    }
    printf("\n");
    return 0;
}