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 c i burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.
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 c 1, c 2, ..., c n (1 ≤ c i ≤ 107), here c i is the cost of delaying the i-th flight for one minute.
The first line must contain the minimum possible total cost of delaying the flights.
The second line must contain n different integers t 1, t 2, ..., t n ( k + 1 ≤ t i ≤ k + n), here t i is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.
4 2 1 10 2
3 6 7 4 5
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.