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)

import itertools


h, w = map(int, input().split())
c = [list(map(ord, input())) for i in range(h)]
c = list(itertools.chain.from_iterable(c))
INF = 10 ** 9


ans = 0
dq = [0] * ((h + 1) ** 2)
for wi in range(w - 1):
    for diff in range(h + 1):
        cnt = 0
        for i in range(h - diff):
            cnt += (c[i * w + wi] == c[(i + diff) * w + wi + 1])
        dq[diff * (h + 1)] = cnt
        for i in range(h - diff - 1):
            cnt -= (c[(h - i - 1 - diff) * w + wi] == c[(h - i - 1) * w + wi + 1])
            dq[(i + 1 + diff) * (h + 1) + i + 1] = cnt

        cnt = 0
        for i in range(h - diff):
            cnt += (c[(i + diff) * w + wi] == c[i * w + wi + 1])
        dq[diff] = cnt
        for i in range(h - diff - 1):
            cnt -= (c[(h - i - 1 - diff) * w + wi + 1] == c[(h - i - 1) * w + wi])
            dq[(i + 1) * (h + 1) + i + 1 + diff] = cnt
            
    dp = [INF] * ((h + 1) ** 2)
    dp[0] = 0
    for i in range(h + 1):
        for j in range(h + 1):
            if i + 1 < h + 1:
                dp[(i + 1) * (h + 1) + j] = min(dp[i * (h + 1) + j] + dq[i * (h + 1) + j], dp[(i + 1) *(h + 1) + j])
            if j + 1 < h + 1:
                dp[i * (h + 1) + j + 1] = min(dp[i * (h + 1) + j] + dq[i * (h + 1) + j], dp[i * (h + 1) + j + 1])
    ans += dp[-1]

print(ans)