Problem:
Score : 800 points

Problem Statement

Mr. Takahashi has in his room an art object with H rows and W columns, made up of H \times W blocks.
Each block has a color represented by a lowercase English letter (a-z).
The color of the block at the i-th row and j-th column is c_{i,j}.

Mr. Takahashi would like to dismantle the object, finding it a bit kitschy for his tastes.
The dismantling is processed by repeating the following operation:

Choose one of the W columns and push down that column one row.
The block at the bottom of that column disappears.

Each time the operation is performed, a cost is incurred.
Since blocks of the same color have a property to draw each other together, the cost of the operation is the number of the pairs of blocks (p, q) such that:

The block p is in the selected column.

The two blocks p and q are horizontally adjacent (before pushing down the column).

The two blocks p and q have the same color.

Mr. Takahashi dismantles the object by repeating the operation H \times W times to get rid of all the blocks.
Compute the minimum total cost to dismantle the object.

Constraints

1 ≤ H ≤ 300

2 ≤ W ≤ 300

All characters c_{i,j} are lowercase English letters (a-z).

Partial Score

In test cases worth 300 points, W = 3.

Input

The input is given from Standard Input in the following format:

H W
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 the minimum total cost to dismantle the object.

Sample Input 1

2 3
rrr
brg

Sample Output 1

2

For example, the total cost of 2 can be achieved by performing the operation as follows and this is the minimum value.

Sample Input 2

6 3
xya
xya
ayz
ayz
xaz
xaz

Sample Output 2

0

The total cost of 0 can be achieved by first pushing down all blocks of the middle column, then all of the left column, and all of the right column.

Sample Input 3

4 2
ay
xa
xy
ay

Sample Output 3

0

The total cost of 0 can be achieved by the following operations:

pushing down the right column one row;

pushing down the left column one row;

pushing down all of the right column;

and pushing down all of the left column.

Sample Input 4

5 5
aaaaa
abbba
ababa
abbba
aaaaa

Sample Output 4

24

Sample Input 5

7 10
xxxxxxxxxx
ccccxxffff
cxxcxxfxxx
cxxxxxffff
cxxcxxfxxx
ccccxxfxxx
xxxxxxxxxx

Sample Output 5

130

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

h, w = STDIN.gets.split(' ').map{|e|e.to_i}
cmap = []
h.times do
  cmap.push(STDIN.gets.split('').map{|e| e.ord})
end

cost = 0
# p cmap
for i in (0...h) do
  for j in (0...w) do
    bottom = cmap[-1]
    bottom_v = [0]*w
    for cx in (0...w) do
      next if bottom[cx] == -1
      for y in (0...h) do
        unless cx == 0 then
          bottom_v[cx] += 1 if bottom[cx] == cmap[y][cx-1]
        end
        unless cx == w-1 then
          bottom_v[cx] += 1 if bottom[cx] == cmap[y][cx+1]
        end
      end
    end
    # p bottom_v
    maxIndex = bottom_v.each_with_index.max[1]
    for y in (0...h) do
      next if cmap[y][maxIndex] == -1
      unless maxIndex == 0 then
        # p [maxIndex, y]
        cost += 1 if cmap[y][maxIndex] == cmap[y][maxIndex-1]
      end
      unless maxIndex == w-1 then
        # p [maxIndex, y]
        cost += 1 if cmap[y][maxIndex] == cmap[y][maxIndex+1]
      end
    end
    for y in (1...h) do
      cmap[y][maxIndex] = cmap[y-1][maxIndex]
    end
    cmap[0][maxIndex] = -1
  end
end

puts cost