Problem:
Score: 500 points

Problem Statement

New AtCoder City has an infinite grid of streets, as follows:

At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.

A straight street, which we will call East-West Main Street, runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.

A straight street, which we will call North-South Main Street, runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.

There are N residential areas in New AtCoder City. The i-th area is located at the intersection with the coordinates (X_i, Y_i) and has a population of P_i. Each citizen in the city lives in one of these areas.

The city currently has only two railroads, stretching infinitely, one along East-West Main Street and the other along North-South Main Street.

M-kun, the mayor, thinks that they are not enough for the commuters, so he decides to choose K streets and build a railroad stretching infinitely along each of those streets.

Let the walking distance of each citizen be the distance from his/her residential area to the nearest railroad.

M-kun wants to build railroads so that the sum of the walking distances of all citizens, S, is minimized.

For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after building railroads?

Constraints

1 \leq N \leq 15

-10 \ 000 \leq X_i \leq 10 \ 000

-10 \ 000 \leq Y_i \leq 10 \ 000

1 \leq P_i \leq 1 \ 000 \ 000

The locations of the N residential areas, (X_i, Y_i), are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N

Output

Print the answer in N+1 lines.

The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value of S after building railroads for the case K = i-1.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

8
2 2 286017
3 1 262355
2 -2 213815
1 -3 224435
-2 -2 136860
-3 -1 239338
-2 2 217647
-1 3 141903

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in Julia (1.4.0) (judge verdict: Time Limit Exceeded)

# %% library
# ----------

# %% library docs
# ---------------

"""
    @collect [cond] ex

Constructs [`Array`](@ref) from lastly evaluated values from a `for` loop block that appears
  first within given `ex` expression.
If the optional `cond` expression is given, iterations  where the `cond` is `false` are
  effectively filtered out.

```julia-repl
julia> @collect isodd(i) for i = 1:3
           println("i = ", i); i
       end
i = 1
i = 3
2-element Array{Int64,1}:
 1
 3
```

See also: [`@generator`](@ref)
"""
macro collect end

"""
    @generator [cond] ex

Constructs [`Base.Generator`](@ref) from lastly evaluated values from a `for` loop block
  that appears first within given `ex` expression.
If the optional `cond` expression is given, iterations where the `cond` is `false` are
  effectively filtered out.

```julia-repl
julia> @generator isodd(i) for i = 1:3
           println("i = ", i); i
       end |> sum
i = 1
i = 3
4
```

See also: [`@collect`](@ref)
"""
macro generator end

"""
    bruteforcesearch(n::Integer, b::Integer = 2)
    bruteforcesearch(f::Function, n::Integer, b::Integer = 2)

$(""#=TODO: add some equation here for the future reference=#)

!!! note
    By default (i.e. when `b == 2`), this function is equivalent to bit-brute-force search.

```julia-repl
julia> for gen in bruteforcesearch(3)
           @show collect(gen)
       end
collect(gen) = [2, 1, 1]
collect(gen) = [1, 2, 1]
collect(gen) = [2, 2, 1]
collect(gen) = [1, 1, 2]
collect(gen) = [2, 1, 2]
collect(gen) = [1, 2, 2]
collect(gen) = [2, 2, 2]
collect(gen) = [1, 1, 1]

julia> bruteforcesearch(2, 3) do gen
           @show collect(gen)
       end |> collect;
collect(gen) = [2, 1]
collect(gen) = [3, 1]
collect(gen) = [1, 2]
collect(gen) = [2, 2]
collect(gen) = [3, 2]
collect(gen) = [1, 3]
collect(gen) = [2, 3]
collect(gen) = [3, 3]
collect(gen) = [1, 1]
```
"""
function bruteforcesearch end

# %% library body
# ---------------

function decompose_forblk(forblk)
    @assert Meta.isexpr(forblk, :for) "for block expression should be given"
    itrspec, body = forblk.args
    @assert Meta.isexpr(itrspec, :(=)) "invalid for loop specification"
    v, itr = itrspec.args
    return body, v, itr
end

function recompose_to_comprehension(forblk, cond = nothing; gen = false)
    body, v, itr = decompose_forblk(forblk)
    return isnothing(cond) ?
        esc(gen ? :(($body for $v in $itr)) : :([$body for $v in $itr])) :
        esc(gen ? :(($body for $v in $itr if $cond)) : :([$body for $v in $itr if $cond]))
end

function walk_and_transform(x, cond = nothing; gen = false)
    Meta.isexpr(x, :for) && return recompose_to_comprehension(x, cond; gen = gen), true
    x isa Expr || return x, false
    for (i, ex) in enumerate(x.args)
        ex, transformed = walk_and_transform(ex, cond; gen = gen)
        x.args[i] = ex
        transformed && return x, true # already transformed
    end
    return x, false
end

macro collect(ex) first(walk_and_transform(ex)) end
macro collect(cond, ex) first(walk_and_transform(ex, cond)) end

macro generator(ex) first(walk_and_transform(ex; gen = true)) end
macro generator(cond, ex) first(walk_and_transform(ex, cond; gen = true)) end

@inbounds begin

function bruteforcesearch(n::Integer, b::Integer = 2)
    baselen = length(string(b, base = 10))
    return @generator for i = 1:b^n
        s = reverse(string(i, pad = n, base = b))[1:n] # cut off the overflowed char when `i == base^n`
        @generator for ns in partition(s, baselen)
            parse(Int, ns) + 1 # for 1-based indexing
        end
    end
end

function bruteforcesearch(f::Function, n::Integer, b::Integer = 2)
    baselen = length(string(b, base = 10))
    return @generator for i = 1:b^n
        s = reverse(string(i, pad = n, base = b))[1:n] # cut off the overflowed char when `i == base^n`
        gen = @generator for ns in partition(s, baselen)
            parse(Int, ns) + 1 # for 1-based indexing
        end
        f(gen)
    end
end

function partition(a, n)
    return if n == 1
        a
    else
        @collect for i in 1:(length(a)÷n)
            s = 1+n*(i-1)
            e = n*i
            a[s:e]
        end
    end
end

end # @inbounds

# %% constants
# ------------

# %% body
# -------

function main(io = stdin)
    readto(target = '\n') = readuntil(io, target)
    readnum(T::Type{<:Number} = Int; dlm = isspace, kwargs...) =
        parse.(T, split(readto(), dlm; kwargs...))

    # handle IO and stuff
    N, = readnum()
    XYPs = [readnum() for _ in 1:N]
    println.(solve(N, XYPs))
end

@inbounds function solve(N, XYPs)
    # TODO: precomputation
    Xcaches = [Dict{UInt,Int}() for _ in 1:N]
    Ycaches = [Dict{UInt,Int}() for _ in 1:N]

    for roads in bruteforcesearch(N, 2)
        Xs = Set(X for (road, (X, _)) in zip(roads, XYPs) if road == 1)
        Ys = Set(Y for (road, (_, Y, _)) in zip(roads, XYPs) if road == 1)
        push!(Xs, 0); push!(Ys, 0)
        cache_distances!(XYPs, Xs, Ys, Xcaches, Ycaches)
    end

    ret = [typemax(Int) for n in 0:N]

    # calculate shortest distances for all the candidates
    for roads in bruteforcesearch(N, 3)
        Xs = Set(0)
        Ys = Set(0)
        cnt = @generator for (road, (X, Y, _)) in zip(roads, XYPs)
            if road == 1
                push!(Xs, X)
                1
            elseif road == 2
                push!(Ys, Y)
                1
            else # no road here, no count
                0
            end
        end |> sum
        ret[cnt + 1] = min(ret[cnt + 1], sumup_distances(XYPs, Xs, Ys, Xcaches, Ycaches))
    end

    return ret
end

function cache_distances!(XYPs, Xs, Ys, Xcaches, Ycaches)
    for ((X, Y, _), Xcache, Ycache) in zip(XYPs, Xcaches, Ycaches)
        cache_distance!(Xcache, X, Xs)
        cache_distance!(Ycache, Y, Ys)
    end
end

function cache_distance!(cache, p, ps)
    k = hash(ps)
    haskey(cache, k) && return
    cache[k] = shortest_distance(p, ps)
end

shortest_distance(p, ps) = minimum(abs(p - x) for x in ps)

@inbounds function sumup_distances(XYPs, Xs, Ys, Xcaches, Ycaches)
    xk = hash(Xs)
    yk = hash(Ys)
    @generator for ((_, _, P), Xcache, Ycache) in zip(XYPs, Xcaches, Ycaches)
        min(Xcache[xk], Ycache[yk]) * P
    end |> sum
end

@static if @isdefined(Juno)
    using BenchmarkTools

    main(open(replace(@__FILE__, r"(.+)\.jl" => s"\1.in")))
    @btime main(open(replace(@__FILE__, r"(.+)\.jl" => s"\1.in")))
else
    main()
end