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)

# 考えると、隣り合う列は結局 2*H 回ずることになり、
# それぞれの列間について最適な沈め順を見つけると、
# その順序関係を崩さないように全体の順番も構築できる（トポロジカルソートみたいに）ので、
# 各列間の最適解を求めて総和をとればいい

# TLE が厳しすぎる

def accumulate(S1, S2):
    res = [0]
    c = 0
    for i in range(min(len(S1), len(S2))):
        if S1[i] == S2[i]:
            c += 1
        res.append(c)
    return res

inf = 10**8
H, W = map(int, input().split())
C = [input() for _ in range(H)]
C_T = list(zip(*C))
C_T = ["".join(s) for s in C_T]
ans = 0
C1 = C_T[0]
for C2 in C_T[1:]:
    # Z[i][j] := i 個ずれてて沈んでない方が j 個沈んでる状態の連結個数
    Z1 = []  # 左を押したとき（右が上にあるとき）
    Z2 = []  # 右を押したとき（左が上にあるとき）
    z1 = accumulate(C1, C2);  z1.reverse()
    Z1.append(z1)
    Z2.append([])
    for j in range(1, H+1):
        z1 = accumulate(C1, C2[j:]);  z1.reverse();  Z1.append(z1)
        z2 = accumulate(C1[j:], C2);  z2.reverse();  Z2.append(z2)
    # dp[i][j] := 左を i 個、右を j 個沈めるまでの最小コスト  # 逆かも（どっちでもいい）
    dp = [[inf] * (H+2) for _ in range(H+2)]
    dp[0][0] = 0
    for j in range(H+1):
        cost = dp[j][0]
        for k in range(H+1):
            cost += (Z1[j-k][k] if j>=k else Z2[k-j][j])
            dp[j+1][k] = min(dp[j+1][k], cost)
            dp[j][k+1] = cost = min(dp[j][k+1], cost)
    ans += dp[H][H]
    C1 = C2
print(ans)