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: Time Limit Exceeded)

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

#pragma warning disable

class Segtree<T>
{
    int n;
    T[] tree;
    Func<T, T, T> f;
    T exnum;
    public Segtree(int m, Func<T, T, T> f, T ex)
    {
        this.f = f;
        this.exnum = ex;
        n = 1;
        while (n < m) n <<= 1;

        tree = new T[(n << 1) - 1];
        for (int i = 0; i < tree.Length; i++) tree[i] = ex;
    }
    public Segtree(int m, T ini, Func<T, T, T> f, T ex)
    {
        this.f = f;
        this.exnum = ex;
        n = 1;
        while (n < m) n <<= 1;

        tree = new T[(n << 1) - 1];
        for (int i = 0; i < tree.Length; i++) tree[i] = ini;
        for (int i = 0; i < m; ++i) update(i, ini);
    }
    public void update(int j, T x)
    {
        int i = j + n - 1;
        tree[i] = x;
        while (i > 0)
        {
            i = (i - 1) >> 1;
            tree[i] = f(tree[(i << 1) + 1], tree[(i << 1) + 2]);
        }
    }
    public T look(int i) { return tree[i + n - 1]; }

    // [s, t]
    public T run(int s, int t) { return query(s, t + 1, 0, 0, n); }
    T query(int s, int t, int k, int l, int r)
    {
        if (r <= s || t <= l) return exnum;
        if (s <= l && r <= t) return tree[k];

        return f(query(s, t, (k << 1) + 1, l, (l + r) >> 1), query(s, t, (k + 1) << 1, (l + r) >> 1, r));
    }
}

class PriorityQueue<T>
{
    private Comparison<T> comp;
    private List<T> list;
    private int count;
    public PriorityQueue() : this((x, y) => ((IComparable<T>)x).CompareTo(y)) { }
    public PriorityQueue(Comparison<T> comparison)
    {
        comp = comparison;
        list = new List<T>();
        count = 0;
    }
    public void Enqueue(T x)
    {
        var pos = count++;
        list.Add(x);
        while (pos > 0)
        {
            var p = (pos - 1) / 2;
            if (comp(list[p], x) <= 0) break;
            list[pos] = list[p];
            pos = p;
        }
        list[pos] = x;
    }
    public T Dequeue()
    {
        var value = list[0];
        var x = list[--count];
        list.RemoveAt(count);
        if (count == 0) return value;
        var pos = 0;
        while (pos * 2 + 1 < count)
        {
            var a = 2 * pos + 1;
            var b = 2 * pos + 2;
            if (b < count && comp(list[b], list[a]) < 0) a = b;
            if (comp(list[a], x) >= 0) break;
            list[pos] = list[a];
            pos = a;
        }
        list[pos] = x;
        return value;
    }
    public T Peek() { return list[0]; }
    public bool Any() { return count > 0; }
    public int Count() { return count; }
}

namespace Codeforces
{
    class Program
    {
        class mymath
        {
            static int Mod = 1000000007;
            public void setMod(int m) { Mod = m; }
            public bool isprime(long a)
            {
                if (a < 2) return false;
                for (long i = 2; i * i <= a; i++) if (a % i == 0) return false;
                return true;
            }
            public bool[] sieve(int n)
            {
                var isp = new bool[n + 1];
                for (int i = 2; i <= n; i++) isp[i] = true;
                for (int i = 2; i * i <= n; i++) if (isp[i]) for (int j = i * i; j <= n; j += i) isp[j] = false;
                return isp;
            }
            public List<int> getprimes(int n)
            {
                var prs = new List<int>();
                var isp = sieve(n);
                for (int i = 2; i <= n; i++) if (isp[i]) prs.Add(i);
                return prs;
            }
            public long[][] E(int n)
            {
                var ret = new long[n][];
                for (int i = 0; i < n; i++)
                {
                    ret[i] = new long[n];
                    ret[i][i] = 1;
                }
                return ret;
            }
            public long[][] powmat(long[][] A, long n)
            {
                if (n == 0) return E(A.Length);
                var t = powmat(A, n / 2);
                if ((n & 1) == 0) return mulmat(t, t);
                return mulmat(mulmat(t, t), A);
            }
            public long[] mulmat(long[][] A, long[] x)
            {
                int n = A.Length, m = x.Length;
                var ans = new long[n];
                for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) ans[i] = (ans[i] + x[j] * A[i][j]) % Mod;
                return ans;
            }
            public long[][] mulmat(long[][] A, long[][] B)
            {
                int n = A.Length, m = B[0].Length, l = B.Length;
                var ans = new long[n][];
                for (int i = 0; i < n; i++)
                {
                    ans[i] = new long[m];
                    for (int j = 0; j < m; j++) for (int k = 0; k < l; k++) ans[i][j] = (ans[i][j] + A[i][k] * B[k][j]) % Mod;
                }
                return ans;
            }
            public long[] addmat(long[] x, long[] y)
            {
                int n = x.Length;
                var ans = new long[n];
                for (int i = 0; i < n; i++) ans[i] = (x[i] + y[i]) % Mod;
                return ans;
            }
            public long[][] addmat(long[][] A, long[][] B)
            {
                int n = A.Length, m = A[0].Length;
                var ans = new long[n][];
                for (int i = 0; i < n; i++) ans[i] = addmat(A[i], B[i]);
                return ans;
            }
            public long powmod(long a, long b)
            {
                if (a >= Mod) return powmod(a % Mod, b);
                if (a == 0) return 0;
                if (b == 0) return 1;
                var t = powmod(a, b / 2);
                if ((b & 1) == 0) return t * t % Mod;
                return t * t % Mod * a % Mod;
            }
            public long inv(long a) { return powmod(a, Mod - 2); }
            public long gcd(long a, long b)
            {
                while (b > 0) { var t = a % b; a = b; b = t; }
                return a;
            }
            public long lcm(long a, long b) { return a * (b / gcd(a, b)); }
            public long Comb(int n, int r)
            {
                if (n < 0 || r < 0 || r > n) return 0;
                if (n - r < r) r = n - r;
                if (r == 0) return 1;
                if (r == 1) return n;
                var numerator = new int[r];
                var denominator = new int[r];
                for (int k = 0; k < r; k++)
                {
                    numerator[k] = n - r + k + 1;
                    denominator[k] = k + 1;
                }
                for (int p = 2; p <= r; p++)
                {
                    int pivot = denominator[p - 1];
                    if (pivot > 1)
                    {
                        int offset = (n - r) % p;
                        for (int k = p - 1; k < r; k += p)
                        {
                            numerator[k - offset] /= pivot;
                            denominator[k] /= pivot;
                        }
                    }
                }
                long result = 1;
                for (int k = 0; k < r; k++) if (numerator[k] > 1) result = result * numerator[k] % Mod;
                return result;
            }
        }

        const long InfL = 4011686018427387913L;
        static char[] sep = new char[] { ' ', '/' };


        //static int[] dx = { -1, 0, 1, -1, 1, -1, 0, -1 };
        //static int[] dy = { 1, 1, 1, 0, 0, -1, -1, -1 };

        static int[] dx = { -1, 0, 0, 1 };
        static int[] dy = { 0, 1, -1, 0 };


        /// <summary>
        /// basic dfs function
        /// </summary>
        /// 
        /* public class List_of_List<T> : List<List<T>> { }
        static int result = 0;
 
        static List<bool> visited = new List<bool>();
        static List_of_List<int> adj = new List_of_List<int>();
 
        static void dfs(int here)
        {
            if (here == n * m - 1)
            {
                result++;
                return;
            }
            visited[here] = true;
 
            for(int i = 0; i<adj[here].Count; i++)
            {
                int there = adj[here][i];
                if (!visited[there])
                    dfs(there);
                visited[there] = false;   
            }
        }
 
        static void dfsAll()
        {
            for(int i = 0; i<adj.Count; i++)
            {
                dfs(i);
            }
        }
 
 
        /* static int getInt => int.Parse(Console.ReadLine());
        static int[] getIntArr => Console.ReadLine().Trim().Split(sep).Select(a => int.Parse(a)).ToArray();
        static List<int> getIntList => Console.ReadLine().Trim().Split(sep).Select(a => int.Parse(a)).ToList<int>();
        static string getString => Console.ReadLine().Trim();
        static string[] getStrArr => Console.ReadLine().Trim().Split(sep);
        static List<string> getStrList => Console.ReadLine().Trim().Split(sep).ToList<string>();
        */
        static int m;
        static int n;
        static int[] input = new int[100001];



        static void Main(string[] args)
        {
            int[] first_arr = Console.ReadLine().Trim().Split(sep).Select(a => int.Parse(a)).ToArray();
            int h = first_arr[0];
            int w = first_arr[1];

            if (w == 2)
            {
                Console.WriteLine(0);
                return;
            }
            int count = h * w;

            int[,] board = new int[h, w];
            for (int i = 0; i < h; i++)
            {
                string s = Console.ReadLine();
                for (int j = 0; j < w; j++)
                {
                    board[i, j] = s[j];
                }
            }

            int result = 0;

            while (true)
            {
                int w1_cost = 0;
                int w2_cost = 0;
                int w3_cost = 0;

                for (int i = 0; i < h; i++)
                {
                    if (board[i, 0] == board[i, 1] && board[i, 0] != 0)
                    {
                        w1_cost++;
                        w2_cost++;
                    }
                    if (board[i, 1] == board[i, 2] && board[i, 2] != 0)
                    {
                        w2_cost++;
                        w3_cost++;
                    }
                }

                if (w2_cost > w1_cost + w3_cost)
                {
                    result += w1_cost + w3_cost;
                    board[h - 1, 0] = 0;
                    board[h - 1, 2] = 0;
                    for (int i = h - 1; i > 0; i--)
                    {
                        board[i, 0] = board[i - 1, 0];
                        board[i, 2] = board[i - 1, 2];
                        board[i - 1, 0] = 0;
                        board[i - 1, 2] = 0;
                    }
                }
                else
                {
                    result += w2_cost;
                    board[h - 1, 1] = 0;
                    for (int i = h - 1; i > 0; i--)
                    {
                        board[i, 1] = board[i - 1, 1];
                        board[i - 1, 1] = 0;
                    }
                }

                if (board[h - 1, 0] == 0 && board[h - 1, 1] == 0 && board[h - 1, 2] == 0) break;
            }
            Console.WriteLine(result);

        }

        /*
        /// <summary>
        /// 이분 매칭 알고리즘
        /// </summary>
 
        static bool[,] adj; // 연결 리스트. 이분 매칭이기 때문에 bool 형식이다. 혹-시라도 타임아웃이 뜬다면 인접리스트로 바꿔볼 것. 조금 빨라진다.
        static List<bool> visited; // 왼쪽의 한 정점에서 오른쪽의 모든 정점으로 방문 기록을 정리하는 리스트
        static List<int> m_match; // 왼쪽에 있는 정점들의 집합
        static List<int> n_match; // 오른쪽에 있는 정점들의 집합
 
 
        static bool dfs(int a) // 왼쪽 정점 a 에서 오른쪽 정점으로 가는 경로를 찾는다
        {
            if (visited[a]) return false;
            visited[a] = true;
            for (int b = 0; b < n; b++)
            {
                if (adj[a, b]) // 왼쪽 정점 a 에서 오른쪽 정점중 b에 연결할 수 있는 경우
                {
                    if (n_match[b] == -1 || dfs(n_match[b])) // b가 이미 매칭되어 있다면 n_match[b]에서 시작되는 증가경로를 찾는다
                    {
                        // 증가 경로를 발견한 경우 a와 b를 매치시키고 true를 반환한다
                        m_match[a] = b;
                        n_match[b] = a;
                        return true;
                    }
                }
            }
            return false;
        }
 
        static int bipartite_match()
        {
            m_match = Enumerable.Repeat(-1, m).ToList(); // 왼쪽에 있는 정점들의 각 원소가 오른쪽에 있는 어떤 정점과 대응되는지 기록하는 리스트
            n_match = Enumerable.Repeat(-1, n).ToList(); // 오른쪽에 있는 정점들의 각 원소가 왼쪽에 있는 어떤 정점과 대응되는지 기록하는 리스트
 
            int size = 0; // 몇 쌍이 대응되는지 기록하는 변수
            for (int start = 0; start < m; start++) // m_match (왼쪽에 있는 정점들 집합) 의 모든 원소에 대해 dfs 실시
            {
                visited = Enumerable.Repeat(false, m).ToList(); // 모든 점에 대해 초기화 (방문하지 않음으로 표시)
 
                if (dfs(start)) size++; // 각 왼쪽 정점에서 dfs를 실시해서 갈 수 있는 경로가 있다면 경로의 개수를 1 늘린다
            }
            return size;
        }
        */

        /*
        /// <summary>
        /// 네트워크 유량 문제용 포드-풀커슨 알고리즘
        /// </summary>
 
        const int INF = 987654321;
        static int v; // 정점의 수. source와 sink가 있으므로 감안해서 계산
        static int max_v;
        static int[,] capacity, flow;
        // capacity[u,v] = u에서 v로 보낼 수 있는 용량, flow[u,v] = u에서 v로 흘러가는 유량 (반대방향인 경우 음수)
 
        // flow[,] 를 계산하고 총 유량을 반환하는 함수
        static int networkFlow(int source, int sink)
        {
            flow = new int[max_v, max_v]; // flow 배열의 초기화
            int totalFlow = 0;
            while (true)
            {
                List<int> parent = Enumerable.Repeat(-1, max_v).ToList();
                Queue<int> q = new Queue<int>();
                parent[source] = source;
                q.Enqueue(source);
                while (q.Count != 0 && parent[sink] == -1)
                {
                    int here = q.Dequeue();
                    for (int there = 0; there < v; there++)
                    {
                        if (capacity[here, there] - flow[here, there] > 0 && parent[there] == -1)
                        {
                            q.Enqueue(there);
                            parent[there] = here;
                        }
                    }
                }
 
                if (parent[sink] == -1) break;
 
                int amount = INF;
                for(int p = sink; p != source; p = parent[p])
                {
                    amount = Math.Min(capacity[parent[p], p] - flow[parent[p], p], amount);
                }
                for(int p = sink; p != source; p = parent[p])
                {
                    flow[parent[p], p] += amount;
                    flow[p, parent[p]] -= amount;
                }
             
                totalFlow += amount;
            }
            return totalFlow;
        }
        */
    }
}