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.IOException;
import java.io.InputStream;
import java.io.PrintStream;
import java.util.NoSuchElementException;

public class Main {
	static FastScanner sc = new FastScanner();
	static PrintStream out = System.out;

	static int H;
	static int W;
	static int c[][];
	public static void main(String args[]) {
		H = ni();
		W = ni();
		c = new int[H][W];
		for (int i=0;i<H;i++) {
			String C = next();
			for (int j=0;j<W;j++) {
				c[i][j] = C.charAt(j) - 'a';
			}
		}

		long dp[][][] = new long[H+1][H+1][H+1];
		for (int i=0;i<H+1;i++) {
			for (int j=0;j<H+1;j++) {
				for (int k=0;k<H+1;k++) {
					dp[i][j][k] = Long.MAX_VALUE;
				}
			}
		}

		dp[0][0][0] = 0;
		for (int i=1;i<=H*W;i++) {
			rec(dp, i, 0, new int[3], 0);
		}

		
		if (W == 2) {
			out.println(dp[H][H][0]);
		} else {
			out.println(dp[H][H][H]);
		}
	}
	
	private static void rec(long dp[][][], int sum, int s, int n[], int d) {
		if (d == W-1) {
			int m = sum - s;
			n[d] = m;

			if (m < 0 || m > H) return;
			
			if (W == 2) n[2] = 0;
			for (int i=0;i<W;i++) {
				if (n[i] == 0) continue;
				long cost = 0;
				long prev = 0;
				if (i == 0) prev = dp[n[0]-1][n[1]][n[2]];
				else if (i == 1) prev = dp[n[0]][n[1]-1][n[2]];
				else if (i == 2) prev = dp[n[0]][n[1]][n[2]-1];

				for (int j=0;j<H-(n[i]-1);j++) {
					if (i > 0) {
						int k = j + ((n[i]-1) - n[i-1]);
						if (0 <= k && k < H) {
							if (c[k][i-1] == c[j][i]) {
								cost++;
							}
						}
					}
					if (i < W-1) {
						int k = j + ((n[i]-1) - n[i+1]);
						if (0 <= k && k < H) {
							if (c[k][i+1] == c[j][i]) {
								cost++;
							}
						}
					}
				}
				
				dp[n[0]][n[1]][n[2]] = Math.min(dp[n[0]][n[1]][n[2]], prev + cost);
			}

			return;
		}
		
		for (int i=0;i<=H;i++) {
			n[d] = i;
			rec(dp, sum, s + i, n, d + 1);
		}
	}

	private static String next() {
		return sc.next();
	}

	private static int ni() {
		return sc.nextInt();
	}

	private static long nl() {
		return sc.nextLong();
	}

	private static int[] na(int n) {
		int[] a = new int[n];
		for(int i = 0;i < n;i++) a[i] = ni();
		return a;
	}

	private static long[] nal(int n) {
		long[] a = new long[n];
		for(int i = 0;i < n;i++) a[i] = ni();
		return a;
	}

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

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
    public boolean hasNext() { skipUnprintable(); return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public int nextInt() {
        if (!hasNext()) throw new NoSuchElementException();
        int n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
}