Problem:
Score : 300 points

Problem Statement

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is ., and black if c_{i,j} is #.

Consider doing the following operation:

Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.

You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Constraints

1 \leq H, W \leq 6

1 \leq K \leq HW

c_{i,j} is . or #.

Input

Input is given from Standard Input in the following format:

H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W}

Output

Print an integer representing the number of choices of rows and columns satisfying the condition.

Sample Input 1

2 3 2
..#
###

Sample Output 1

5

Five choices below satisfy the condition.

The 1-st row and 1-st column

The 1-st row and 2-nd column

The 1-st row and 3-rd column

The 1-st and 2-nd column

The 3-rd column

Sample Input 2

2 3 4
..#
###

Sample Output 2

1

One choice, which is choosing nothing, satisfies the condition.

Sample Input 3

2 2 3
##
##

Sample Output 3

0

Sample Input 4

6 6 8
..##..
.#..#.
#....#
######
#....#
#....#

Sample Output 4

208

Solution in Ruby (2.7.1) (judge verdict: Accepted)

h, w , kkk = gets.chomp.split(" ").map(&:to_i)
arr = []
h.times do 
    a = gets.chomp.split("")
    arr << a
end
max = []
["o","x"].repeated_permutation (h) do |i|
    hairetsu = []
    i.each_with_index do |ii,bangou|
        if ii == "o"
             arr_size = arr[bangou].size
             ar = []
             arr_size.times do
                ar << "%"
             end
        hairetsu  << ar
            
        end
        if ii == "x"
            hairetsu << arr[bangou]
        end
    end
    max << hairetsu
end

count = 0
# max.each do |k|
#     ["o","x"].repeated_permutation (w) do |i|
#         i.each_with_index do |ii,bangou|
#             if ii == "o"
#                 h.times do |l|
#                      k[l][bangou].fill("%")
#                 end
#            end
#         end
#         p k
#         kk = k.count(".")
#         if kk == kkk
#             count += 1
#         end
#     end
# end

max.each do |k|
   
    max = []
    k = k.transpose
    ["o","x"].repeated_permutation (w) do |i|
        ookisa = 0
        hairetsu = []
        i.each_with_index do |ii,bangou|
            if ii == "o"
                arr_size = k[bangou].size
                ar = []
                arr_size.times do
                    ar << "%"
                end
            hairetsu  << ar
                
            end
            if ii == "x"
                hairetsu << k[bangou]
            end
        end
        max << hairetsu
        max = max.flatten
        #p max
        ookisa = max.count("#")
        max = []
        if ookisa == kkk
            count += 1
        end
    end
end
puts count