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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

n,m = map(int,input().split())
a = [input() for i in range(n)]
b = list(zip(*a))
for i in range(m):
  b[i] = "".join(b[i])
ans = 0
for i in range(m-1):
  b1 = b[i]
  b2 = b[i+1]
  ct1 = [[0 for k in range(n-j+1)] for j in range(n+1)]
  for j in range(n+1):
    b2c = b2[:n-j]
    for k in range(n-j):
      if b1[k+j] == b2c[k]:
        ct1[j][k+1] += ct1[j][k]+1
      else:
        ct1[j][k+1] += ct1[j][k]
  ct2 = [[0 for k in range(n-j+1)] for j in range(n+1)]
  for j in range(n+1):
    b1c = b1[:n-j]
    for k in range(n-j):
      if b2[k+j] == b1c[k]:
        ct2[j][k+1] += ct2[j][k]+1
      else:
        ct2[j][k+1] += ct2[j][k]
  dp = [[10**6 for i in range(n+1)] for j in range(n+1)]
  dp[0][0] = 0
  for i in range(n+1):
    for j in range(n+1):
      if i <= j:
        cost = ct2[j-i][n-j]
      else:
        cost = ct1[i-j][n-i]
      if i < n:
        dp[i+1][j] = min(dp[i+1][j],dp[i][j]+cost)
      if j < n:
        dp[i][j+1] = min(dp[i][j+1],dp[i][j]+cost)
  ans += dp[-1][-1]
print(ans)