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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
using System.Threading;

using Pair = System.Collections.Generic.KeyValuePair<int, int>;

class Program
{
	public Program() { }

	static void Main(string[] args)
	{
		new Program().prog();
	}
	Scanner cin;
	const int MOD = 1000000007;
	const int INF = int.MaxValue;
	const double EPS = 1e-7;
	const double PI = 3.1415926536;

	void prog()
	{
		cin = new Scanner();
		int[,] dir8 = new int[8, 2] { { -1, -1 }, { -1, 0 }, { -1, 1 }, { 0, -1 }, { 0, 1 }, { 1, -1 }, { 1, 0 }, { 1, 1 } };
		int[,] dir4 = new int[4, 2] { { -1, 0 }, { 0, -1 }, { 0, 1 }, { 1, 0 } };

		int H = cin.nextInt();
		int W = cin.nextInt();
		char[,] c1 = new char[H, W];
		char[,] c = new char[W, H];
		long ans = 0;
		for (int i = 0; i < H; i++)
		{
			string str = cin.next();
			for (int j = 0; j < W; j++)
			{
				c1[i, j] = str[j];
			}
		}
		for (int i = 0; i < W; i++)
		{
			for (int j = 0; j < H; j++)
			{
				c[i, j] = c1[H - 1 - j, i];
			}
		}

		if (W > 3) return;

		int[,] cost = new int[H + 1, H + 1];
		for (int i = 0; i < W - 1; i++)
		{
			// 列i, i+1の間に発生するコスト

			for (int j = 0; j <= H; j++)
			{
				cost[H, j] = 0;
				for (int k = 0; k < j; k++)
				{
					if (c[i, k] == c[i + 1, k + H - j]) cost[H, j]++;
				}
			}
			for (int j = 0; j <= H; j++)
			{
				cost[j, H] = 0;
				for (int k = 0; k < j; k++)
				{
					if (c[i, k + H - j] == c[i + 1, k]) cost[j, H]++;
				}
			}
			for (int j = H - 1; j >= 0; j--)
			{
				for (int k = H - 1; k >= 0; k--)
				{
					cost[j, k] = cost[j + 1, k + 1] - (c[i, H - j - 1] == c[i + 1, H - k - 1] ? 1 : 0);
				}
			}

			for (int j = 1; j <= H; j++)
			{
				for (int k = 1; k <= H; k++)
				{
					cost[j, k] += Math.Min(cost[j - 1, k], cost[j, k - 1]);
				}
			}
			ans += cost[H, H];
		}
		Console.WriteLine(ans);
	}
}

class Scanner
{
	string[] s;
	int i;

	char[] cs = new char[] { ' ' };

	public Scanner()
	{
		s = new string[0];
		i = 0;
	}

	public string next()
	{
		if (i < s.Length) return s[i++];
		string st = Console.ReadLine();
		while (st == "") st = Console.ReadLine();
		s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
		i = 0;
		return next();
	}

	public int nextInt()
	{
		return int.Parse(next());
	}

	public long nextLong()
	{
		return long.Parse(next());
	}

	public double nextDouble()
	{
		return double.Parse(next());
	}
}