IZBORI - IZBORI

It is election time. V voters attend the election, each casting their vote for one of N political parties. M officials will be elected into the parliament.

The conversion from votes to parliament seats is done using the D'Hondt method with a 5% threshold. More precisely, suppose that the parties are numbered 1 through N and that they receive V1, V2, ..., VN votes. Parliament seats are allocated as follows:

  1. All parties that receive strictly less than 5% of V votes are erased from the list of parties.
  2. The parliament is initially empty i.e. every party has zero seats allocated.
  3. For each party P, the quotient QP=VP/(SP+1) is calculated, where VP is the total number of votes received by party P, and SP is the number of seats already allocated to party P.
  4. The party with the largest quotient QP is allocated one seat. If multiple parties have the same largest quotient, the lower numbered party wins the seat.
  5. Repeat steps 3 and 4 until the parliament is full. The votes are being counted and only part of the V votes has been tallied. It is known how many votes each party has received so far.

Write a program that calculates for each party, among all possible outcomes of the election after all V votes are counted, the largest and smallest number of seats the party wins.

Input

The first line contains the integers V, N and M (1 ≤ V ≤ 10,000,000, 1 ≤ N ≤ 100, 1 ≤ M ≤ 200), the numbers of votes, parties and seats in the parliament.

The second line contains N integers – how many votes (of those that have been counted) each party got. The sum of these numbers will be at most V.

Output

On the first line output N integers separated by spaces – the largest number of seats each party can win.

On the second line output N integers separated by spaces – the smallest number of seats each party can win.

Example

Input:
20 4 5
4 3 6 1

Output:
3 3 3 2
1 0 1 0
Input:
100 3 5
30 20 10

Output:
4 3 3
1 1 0

In the first example, 14 votes have been tallied and 6 are yet to be counted. To illustrate one possible outcome, suppose that the first party receives 2 of those 6 votes, the second none, the third 1 vote and the fourth 3 votes. The parties' totals are 6, 3, 7 and 4 votes. All parties exceeded the 5% threshold. Seats are allocated as follows:

  1. The quotients are initially 6/1, 3/1, 7/1 and 4/1; the largest is 7/1 so party 3 wins a seat.
  2. The quotients are 6/1, 3/1, 7/2 and 4/1; the largest is 6/1 so party 1 wins a seat.
  3. The quotients are 6/2, 3/1, 7/2 and 4/1; the largest is 4/1 so party 4 wins a seat.
  4. The quotients are 6/2, 3/1, 7/2 and 4/2; the largest is 7/2 so party 3 wins a seat.
  5. The quotients are 6/2, 3/1, 7/3 and 4/2; parties 1 and 2 are tied with quotients 6/2 and 3/1, but party 1 is lower numbered so it wins the last seat.

In this outcome, the numbers of seats won by the parties are 2, 0, 2 and 1. Since it is possible for the second party not to win any seats, the second number on the second line of output is zero.


Được gửi lên bởi:Race with time
Ngày:2009-03-29
Thời gian chạy:0.104s
Giới hạn mã nguồn:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Ngôn ngữ cho phép:Tất cả ngoại trừ: ERL GOSU JS-RHINO NODEJS PERL6 PYPY RUST SED VB.NET
Nguồn bài:Croatian Olympiad in Informatics 28.03.2009.

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