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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.*;
import java.util.*;

class Main{
	int H;
	int W;
	public void run(){
		H = ni();
		W = ni();
		char[][] ob = new char[H][W];
		for(int i=0;i<H;i++){
			String s = next();
			for(int j=0;j<W;j++){
				ob[i][j] = s.charAt(j);
			}
		}
//		debug(ob);
		int[] rm = new int[W];
		out.println(solve(ob, rm, 0));
	}

	int solve(char[][] ob, int[] rm, int n){
		if(n == (H * W)){
			return 0;
		}
		int min = Integer.MAX_VALUE;
		for(int i=0;i<W;i++){
			if(rm[i] < H){
				//消した時のスコア計算
				//左列
				int score = 0;
				if(i>= 1){
					for(int j=Math.max(rm[i], rm[i-1]);j<H;j++){
//						debug(ob[j-rm[i]][i], ob[j-rm[i-1]][i-1]);
						if(ob[j-rm[i]][i]  == ob[j-rm[i-1]][i-1]){
							score++;
						}
					}
				}
//				debug("\n");
				if(i< W-1){
					for(int j=Math.max(rm[i], rm[i+1]);j<H;j++){
//						debug(ob[j-rm[i]][i], ob[j-rm[i+1]][i+1]);
						if(ob[j-rm[i]][i] == ob[j-rm[i+1]][i+1]){
							score++;
						}
					}
				}
				//右列
				rm[i]++;
				score += solve(ob, rm, n+1);
//				out.println(score);
				if(min > score)min = score;
				rm[i]--;
			}
		}
		return min;
	}

	private static PrintWriter out;
	public static void main(String[] args){
		out = new PrintWriter(System.out);
		new Main().run();
		out.flush();
	}

	public Main(){
	}

	int ni(){
		int num = 0;
		String str = next();
		boolean minus = false;
		int i = 0;
		if(str.charAt(0) == '-'){
			minus = true;
			i++;
		}
		int len = str.length();
		for(;i < len; i++){
			char c = str.charAt(i);
			if(!('0' <= c && c <= '9')) throw new RuntimeException();
			num = num * 10 + (c - '0');
		}
		return minus ? -num : num;	}

	long nl(){
		long num = 0;
		String str = next();
		boolean minus = false;
		int i = 0;
		if(str.charAt(0) == '-'){
			minus = true;
			i++;
		}
		int len = str.length();
		for(;i < len; i++){
			char c = str.charAt(i);
			if(!('0' <= c && c <= '9')) throw new RuntimeException();
			num = num * 10l + (c - '0');
		}
		return minus ? -num : num;
	}

	double nd(){
		return Double.parseDouble(next());
	}

	String next(){
		int c;
		while(!isAlNum(c = read()));
		StringBuilder build = new StringBuilder();
		build.append((char)c);
		while(isAlNum(c = read())){
			build.append((char)c);
		}
		return build.toString();
	}
	String nextLine(){
		int c;
		while(!isAlNumOrSpace(c = read()));
		StringBuilder build = new StringBuilder();
		build.append((char)c);
		while(isAlNumOrSpace(c = read())){
			build.append((char)c);
		}
		return build.toString();
	}
	private static byte[] inputBuffer = new byte[1024];
	private static int bufferLength = 0;
	private static int bufferIndex = 0;
	
	private static int read(){
		if(bufferLength < 0) throw new RuntimeException();
		if(bufferIndex >= bufferLength){
			try{
				bufferLength = System.in.read(inputBuffer);
				bufferIndex = 0;
			}catch(IOException e){
				throw new RuntimeException(e);
			}
			if(bufferLength <= 0) return (bufferLength = -1);
		}
		return inputBuffer[bufferIndex++];
	}
	
	private static boolean isAlNum(int c){
		return '!' <= c && c <= '~';
	}
	private static boolean isAlNumOrSpace(int c){
		return isAlNum(c) || c == ' ' || c == '\t';
	}
	void debug(Object... os){
		out.println(Arrays.deepToString(os));
	}
}