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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

H, W = list(map(int, input().split()))
X = [list(map(ord,input())) for _ in range(H)]
Y = list(map(list, zip(*X)))


def get(A, B):
    cost = [[0 for _ in range(H + 2)] for _ in range(H + 2)]

    def calc(A, B, a_n, b_n):

        if a_n >= H or b_n >= H:
            return 0

        return cost[a_n][b_n]

    def set_cost(A, B, a_n, b_n):
        start = max(a_n, b_n)
        ans = 0
        for i in range(start, H):
            if A[i - a_n] == B[i - b_n]:
                ans += 1
            cost[a_n + H - 1 - i][b_n + H - 1 - i] = ans #cache

    for i in range(H):
        set_cost(A,B,0,i)
        set_cost(A,B,i,0)

    dp = [[999999999999999 for _ in range(H + 2)] for _ in range(H + 2)]
    dp[0][0] = 0

    for i in range(H + 1):
        for j in range(H + 1):
            dp[i + 1][j] = min(dp[i][j] + calc(A, B, i, j), dp[i + 1][j])
            dp[i][j + 1] = min(dp[i][j] + calc(A, B, i, j), dp[i][j + 1])
    # print(dp)

    return dp[H][H]


ans = 0
for i in range(W - 1):
    tmp = get(Y[i], Y[i + 1])
    ans += tmp
    # print(tmp)
print(ans)