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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

type bufReader struct {
	r   *bufio.Reader
	buf []byte
	i   int
}

var reader = &bufReader{
	bufio.NewReader(os.Stdin),
	make([]byte, 0),
	0,
}

func (r *bufReader) readLine() {
	if r.i < len(r.buf) {
		return
	}
	r.buf = make([]byte, 0)
	r.i = 0
	for {
		line, isPrefix, err := r.r.ReadLine()
		if err != nil {
			panic(err)
		}
		r.buf = append(r.buf, line...)
		if !isPrefix {
			break
		}
	}
}

func (r *bufReader) next() string {
	r.readLine()
	from := r.i
	for ; r.i < len(r.buf); r.i++ {
		if r.buf[r.i] == ' ' {
			break
		}
	}
	s := string(r.buf[from:r.i])
	r.i++
	return s
}

func (r *bufReader) nextLine() string {
	r.readLine()
	s := string(r.buf[r.i:])
	r.i = len(r.buf)
	return s
}

var writer = bufio.NewWriter(os.Stdout)

func next() string {
	return reader.next()
}

func nextInt64() int64 {
	i, err := strconv.ParseInt(reader.next(), 10, 64)
	if err != nil {
		panic(err)
	}
	return i
}

func nextInt() int {
	return int(nextInt64())
}

func nextLine() string {
	return reader.nextLine()
}

func out(a ...interface{}) {
	fmt.Fprintln(writer, a...)
}

func max64(x, y int64) int64 {
	if x > y {
		return x
	}
	return y
}

func max(x, y int) int {
	return int(max64(int64(x), int64(y)))
}

func min64(x, y int64) int64 {
	if x < y {
		return x
	}
	return y
}

func min(x, y int) int {
	return int(min64(int64(x), int64(y)))
}

func abs64(x int64) int64 {
	if x < 0 {
		return -x
	}
	return x
}

func abs(x int) int {
	return int(abs64(int64(x)))
}

func joinInt64s(a []int64, sep string) string {
	b := make([]string, len(a))
	for i, v := range a {
		b[i] = strconv.FormatInt(v, 10)
	}
	return strings.Join(b, sep)
}

func joinInts(a []int, sep string) string {
	b := make([]string, len(a))
	for i, v := range a {
		b[i] = strconv.Itoa(v)
	}
	return strings.Join(b, sep)
}

func divUp64(x, y int64) int64 {
	return (x + y - 1) / y
}

func divUp(x, y int) int {
	return int(divUp64(int64(x), int64(y)))
}

func gcd64(x, y int64) int64 {
	if x < y {
		x, y = y, x
	}
	for y != 0 {
		x, y = y, x%y
	}
	return x
}

func gcd(x, y int) int {
	return int(gcd64(int64(x), int64(y)))
}

func lcm64(x, y int64) int64 {
	return x * y / gcd64(x, y)
}

func lcm(x, y int) int {
	return int(lcm64(int64(x), int64(y)))
}

func main() {
	solve()
	writer.Flush()
}

func solve() {
	H, W := nextInt(), nextInt()
	c := make([][]string, H)
	for h := 0; h < H; h++ {
		c[h] = strings.Split(nextLine(), "")
	}
	idxs := make([]int, W)
	for w := 0; w < W; w++ {
		idxs[w] = H - 1
	}
	ans := 0
	for {
		selectedW := -1
		minCost := -1
		for w := 0; w < W; w++ {
			if idxs[w] < 0 || (W == 3 && w != 1 && idxs[1] >= 0) {
				continue
			}
			cost := 0
			for di := 0; di <= idxs[w]; di++ {
				if w > 0 && idxs[w-1]-di >= 0 && c[idxs[w]-di][w] == c[idxs[w-1]-di][w-1] {
					cost++
				}
				if w < W-1 && idxs[w+1]-di >= 0 && c[idxs[w]-di][w] == c[idxs[w+1]-di][w+1] {
					cost++
				}
			}
			if minCost == -1 || cost < minCost {
				selectedW = w
				minCost = cost
			}
		}
		if selectedW == -1 {
			break
		}
		ans += minCost
		idxs[selectedW]--
	}
	out(ans)
}