Problem:
Score : 400 points

Problem Statement

Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left.
Each square is painted black or white.

The grid is said to be good if and only if the following condition is satisfied:

From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.

Note that (1, 1) and (H, W) must be white if the grid is good.

Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.

Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.

Constraints

2 \leq H, W \leq 100

Input

Input is given from Standard Input in the following format:

H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}

Here s_{rc} represents the color of (r, c) - # stands for black, and . stands for white.

Output

Print the minimum number of operations needed.

Sample Input 1

3 3
.##
.#.
##.

Sample Output 1

1

Do the operation with (r_0, c_0, r_1, c_1) = (2, 2, 2, 2) to change just the color of (2, 2), and we are done.

Sample Input 2

2 2
#.
.#

Sample Output 2

2

Sample Input 3

4 4
..##
#...
###.
###.

Sample Output 3

0

No operation may be needed.

Sample Input 4

5 5
.#.#.
#.#.#
.#.#.
#.#.#
.#.#.

Sample Output 4

4

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

=begin
def rec_route (hw_map,r,c,target_h,target_w)

    if r > target_h || c > target_h then
        return target_h*target_h
    end

    if r==target_h && c==target_w && hw_map[r-1][c-1]=="#" then
        return 1
    end
    if r==target_h && c==target_w && hw_map[r-1][c-1]=="." then
        return 0
    end

    flip = 0
    if hw_map[r-1][c-1]=="#" then
        flip = 1
    end

    return [rec_route(hw_map,r,c+1,target_h,target_w), rec_route(hw_map,r+1,c,target_h,target_w)].min + flip

end
=end

hw = gets.split(' ').map(&:to_i)
h = hw[0]
w = hw[1]

src = readlines.map(&:chomp)

hw_map = Array.new(h)
cost_map = Array.new(h)
for i in 0..h-1 do
    hw_map[i] = Array.new(w)
    cost_map[i] = Array.new(w)
end

for r in 0..h-1 do
    for c in 0..w-1 do
        hw_map[r][c] = src[r][c]
        cost_map[r][c] = h*w
    end
end

(0..h-1).reverse_each do |r|
    (0..w-1).reverse_each do |c|
        if r == h-1 && c == w-1 then
            if hw_map[r][c] == "#" then
                cost_map[r][c] = 1
            else
                cost_map[r][c] = 0
            end
        else

            if r+1 > h-1 then
                row_cost = h*w
                cost = h*w
            else
                row_cost = cost_map[r+1][c]
                if hw_map[r+1][c] == "#" then
                    cost = 0
                else
                    cost = 1
                end
            end
            
            if c+1 > w-1 then
                column_cost = h*w
                cost = h*w
            else
                column_cost = cost_map[r][c+1]
                if hw_map[r][c+1] == "#" then
                    cost = 0
                else
                    cost = 1
                end
            end
            min_cost = [row_cost, column_cost].min
            ##低かった方の、前が#ならコストにならない
            cost_map[r][c] = min_cost + cost
        end
    end
end

puts cost_map[0][0]

#ルートを全てだし、黒い色の数が一番少ないのを選ぶ
###puts rec_route(hw_map,1,1,h,w)