Codeforces Round #433 (Div. 2) Planning (贪心+并查集)

 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.

一道明显的贪心题目，优先处理推迟耗费大的航班，使它尽量耗费较少，这样可使总的耗费较少！但是普通遍历明显会TLE，所以我们可以用并查集跳过已经被占用的时间，记录每一个点的下一个点是谁即可。

具体细节代码里说：

#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<cstring>#include<string>#include<vector>#include<map>#include<cmath>using namespace std;#define mem(a,b) memset(a,b,sizeof(a))typedef long long ll;const int maxn = 3e5+5;const int ff = 0x3f3f3f3f;struct node{ll c;int pos;}node[maxn];int n,k; int ans[maxn];int ne[maxn*2];//这个数组要开大bool cmp(struct node a,struct node b){return a.c> b.c;}void init(){for(int i = 0;i<= maxn*2;i++)ne[i] = i;return ;}int find(int x){int r = x;while(x!= ne[x])x = ne[x];while(r!= ne[r])//一定要路径压缩，否则很容易超时{int t = ne[r];ne[r] = x;r = t;}return x;}int main(){cin>>n>>k;init();for(int i = 1;i<= n;i++){scanf("%I64d",&node[i].c);node[i].pos = i;}ll sum = 0;sort(node+1,node+n+1,cmp);for(int i = 1;i<= n;i++){int fa = find(max(node[i].pos,k+1));ans[node[i].pos] = fa;sum+= (fa-node[i].pos)*node[i].c;//运算过程中可能会超int，这里注意ne[fa] = find(fa+1);}cout<<sum<<endl;for(int i = 1;i<= n;i++)if(i == n)printf("%d\n",ans[i]);elseprintf("%d ",ans[i]);return 0;}