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)

H , W = map(int,input().split())
table = [input() for _ in range(H)]
def memo_rec(a,b,i,j, memo):
    if a == H or b == H:
        return 0
    if memo[a][b] > -1:
        return memo[a][b]
    ret = 0
    tmp = 0
    memo[a][b] = min(memo_rec(a+1,b,i,j, memo= memo), memo_rec(a,b+1,i,j,memo=memo)) + cache[a][b]
#    print("(i,j) = (%d,%d) memo[%d][%d] = %d"% (i,j,a,b,memo[a][b]))
    return memo[a][b]
ans = 0
for i in range(W-1):
#    print("%d,%d"%(i,i+1))
    memo = [[-1 for _ in range(H)] for _ in range(H)]
    cache = [[-1 for _ in range(H+1)] for _ in range(H+1)]
    for a in range(H,-1,-1):
        for b in range(H,-1,-1):
            if a == H or b == H:
                cache[a][b] = 0
            elif table[H-a-1][i] == table[H-b-1][i+1]:
                cache[a][b] = cache[a+1][b+1] + 1
            else :
                cache[a][b] = cache[a+1][b+1]
    ans += (memo_rec(0,0,i,i+1,memo))
print (ans)