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: Accepted)

using System;
using System.Collections.Generic;
using System.Collections;
using System.Collections.Specialized;
using System.Linq;
using System.Text;
using System.IO;
using System.Reflection;
using static System.Math;
using System.Numerics;
static class Program{
	const int mod=(int)1e9+7;
	static void Main(){
		Sc sc=new Sc();
		var s=sc.Ia;
		var t=sc.Arr(s[0],()=>sc.S);
		var h=new int[s[1]-1][][];
		var dp=new int[s[1]-1][][];
		for(int i = 0;i<s[1]-1;i++) {
			h[i]=new int[s[0]+1][];
			dp[i]=new int[s[0]+1][];
			for(int j = 0;j<s[0]+1;j++) {
				h[i][j]=new int[s[0]+1];
				dp[i][j]=new int[s[0]+1];
			}
			for(int j = 0;j<s[0];j++) {
				for(int k = s[0]-1;k-j>=0;k--) {
					h[i][k][k-j]=h[i][k+1][k-j+1]+(t[s[0]-k-1][i]==t[s[0]-k+j-1][i+1]?1:0);
					if(j==0){continue;}
					h[i][k-j][k]=h[i][k-j+1][k+1]+(t[s[0]-k+j-1][i]==t[s[0]-k-1][i+1]?1:0);
				}
			}
		}
		long ans=0;
		for(int i = 0;i<s[1]-1;i++) {
			for(int j = 0;j<=s[0];j++) {
				for(int k = 0;k<=s[0];k++) {
					if(j+k>0){dp[i][j][k]=int.MaxValue;}
					if(j>0){dp[i][j][k]=dp[i][j-1][k]+h[i][j-1][k];}
					if(k>0){dp[i][j][k]=Min(dp[i][j][k-1]+h[i][j][k-1],dp[i][j][k]);}
				}
			}
			ans+=dp[i][s[0]][s[0]];
		}
		Console.WriteLine("{0}",ans);
	}
}
public class Sc{
	public int I{get{return int.Parse(Console.ReadLine());}}
	public long L{get{return long.Parse(Console.ReadLine());}}
	public double D{get{return double.Parse(Console.ReadLine());}}
	public string S{get{return Console.ReadLine();}}
	public int[] Ia{get{return Array.ConvertAll(Console.ReadLine().Split(),int.Parse);}}
	public long[] La{get{return Array.ConvertAll(Console.ReadLine().Split(),long.Parse);}}
	public double[] Da{get{return Array.ConvertAll(Console.ReadLine().Split(),double.Parse);}}
	public string[] Sa{get{return Console.ReadLine().Split();}}
	public object[] Oa{get{return Console.ReadLine().Split();}}
	public int[] Ia2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),int.Parse);}}
	public int[] Ia3(int a){return Array.ConvertAll((a.ToString()+" "+Console.ReadLine()).Split(),int.Parse);}
	public int[] Ia3(bool a,int b,bool c,int d){return Array.ConvertAll(((a?b.ToString()+" ":"")+Console.ReadLine()+(c?" "+d.ToString():"")).Split(),int.Parse);}
	public long[] La2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),long.Parse);}}
	public long[] La3(int a){return Array.ConvertAll((a.ToString()+" "+Console.ReadLine()).Split(),long.Parse);}
	public long[] La3(bool a,int b,bool c,int d){return Array.ConvertAll(((a?b.ToString()+" ":"")+Console.ReadLine()+(c?" "+d.ToString():"")).Split(),long.Parse);}
	public double[] Da2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),double.Parse);}}
	public double[] Da3(int a){return Array.ConvertAll((a.ToString()+" "+Console.ReadLine()).Split(),double.Parse);}
	public double[] Da3(bool a,int b,bool c,int d){return Array.ConvertAll(((a?b.ToString()+" ":"")+Console.ReadLine()+(c?" "+d.ToString():"")).Split(),double.Parse);}
	public T[] Arr<T>(int n,Func<T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f();}return a;}
	public T[] Arr<T>(int n,Func<int,T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i);}return a;}
	public T[] Arr<T>(int n,Func<string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(Console.ReadLine().Split());}return a;}
	public T[] Arr<T>(int n,Func<int,string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i,Console.ReadLine().Split());}return a;}
}