// Original C++ solution (slower):
#include <bits/stdc++.h>

//#include <boost/functional/hash.hpp>

//example: unordered_set< pair<int,int>,boost::hash< std::pair<int, int> > > used;

//priority_queue< pair<int,pair<int,int> >, vector<pair<int,pair<int,int>>>, greater<pair<int,pair<int,int> > > > pqueue;    //cost, vertex(行き先)

using namespace std;



#define MODULE 1000000007

#define MP make_pair

#define ALL(v) (v).begin(),(v).end()

#define fi first

#define se second



template<class T, class U>

inline void chmin(T &t, U f) { if (t > f)t = f; }



template<class T, class U>

inline void chmax(T &t, U f) { if (t < f)t = f; }



typedef pair<int, int> P;

typedef long long LL;

const LL INF = LLONG_MAX / 2;    //int_max->2*e+9 LLの時はLLONG_MAX

const int MAXN = 100001;



/*struct edge {

    edge(int to, int cost) : to(to), cost(cost) {}

    int to, cost;

};

vector<edge> graph[MAXN];



int dx[4] = {-1, 0, 1, 0};

int dy[4] = {0, -1, 0, 1};*/



//-----Template---------

int h, w;

vector<string> color(301);

LL sum = 0;

int main() {

    ios::sync_with_stdio(false); //cout<< fixed << setprecision(10);

    cin >> h >> w;

    for (int i = 0; i < h; ++i) {

        cin >> color[i];

    }

    for (int i = 0; i < w - 1; ++i) {     //2列毎の最小コストを求める

        vector<vector<LL>> cost(301,vector<LL>(301));

        for (int j = 0; j <= h; ++j) {

            for (int k = 0; k <= h; ++k) {

                if (j == 0 || k == 0) {

                    LL tmp = 0;

                    for (int l = 0; l < h; ++l) {//0 .. h-1まで回して，変なとこを飛ばす

                        if (l - j >= 0 && l - k >= 0 && color[l - j][i] == color[l - k][i + 1])

                            tmp++;

                    }

                    cost[j][k] = tmp;

                } else {

                    cost[j][k] = cost[j - 1][k - 1] - (color[h - j][i] == color[h - k][i + 1] ? 1 : 0);

                }

            }

        }

        vector<vector<LL>> dp(301,vector<LL>(301,INF));

        dp[0][0] = 0;

        for (int j = 0; j <= h; ++j) {

            for (int k = 0; k <= h; ++k) {

                LL tmpA = j > 0 ? dp[j - 1][k] + cost[j - 1][k] : INF;

                LL tmpB = k > 0 ? dp[j][k - 1] + cost[j][k - 1] : INF;

                chmin(dp[j][k], min(tmpA, tmpB));

            }

        }

        sum += dp[h][h];

    }

    cout << sum << endl;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

//#include <boost/functional/hash.hpp>

//example: unordered_set< pair<int,int>,boost::hash< std::pair<int, int> > > used;

//priority_queue< pair<int,pair<int,int> >, vector<pair<int,pair<int,int>>>, greater<pair<int,pair<int,int> > > > pqueue;    //cost, vertex(行き先)

using namespace std;



#define MODULE 1000000007

#define MP make_pair

#define ALL(v) (v).begin(),(v).end()

#define fi first

#define se second



template<class T, class U>

inline void chmin(T &t, U f) { if (t > f)t = f; }



template<class T, class U>

inline void chmax(T &t, U f) { if (t < f)t = f; }



typedef pair<int, int> P;

typedef long long LL;

const LL INF = LLONG_MAX / 2;    //int_max->2*e+9 LLの時はLLONG_MAX

const int MAXN = 100001;



template<typename A, size_t N, typename T>

inline void Fill(A (&array)[N], const T &val){ //usage: int dp[10][10]; Fill(dp,INF);

    std::fill( (T*)array, (T*)(array+N), val );

}

/*struct edge {

    edge(int to, int cost) : to(to), cost(cost) {}

    int to, cost;

};

vector<edge> graph[MAXN];



int dx[4] = {-1, 0, 1, 0};

int dy[4] = {0, -1, 0, 1};*/



//-----Template---------

int h, w;

char color[301][301];   //vector<string>でもいい

LL dp[301][301];

LL cost[301][301];

LL ans[301];

LL sum = 0;

string tmpS;



int main() {

    ios::sync_with_stdio(false); //cout<< fixed << setprecision(10);

    cin >> h >> w;

    for (int i = 0; i < h; ++i) {

        cin >> tmpS;

        for (int j = 0; j < w; ++j) {

            color[i][j] = tmpS[j];

        }

    }





    for (int i = 0; i < w - 1; ++i) {     //2列毎の最小コストを求める

        for (int j = 0; j <= h; ++j) {   //先にcostテーブルを作っておく．

            // 毎回計算するとdp更新にO(h)かかってTLE

            //cost[i][j]:i,j押されているときに

            //どちらかの列を押すコスト(2列の内どちらを押しても同じコスト)

            for (int k = 0; k <= h; ++k) {

                if (j == 0 || k == 0) {

                    LL tmp = 0;

                    for (int l = 0; l < h; ++l) {//0 .. h-1まで回して，変なとこを飛ばす

                        if (l - j >= 0 && l - k >= 0 && color[l - j][i] == color[l - k][i + 1])

                            tmp++;

                    }

                    cost[j][k] = tmp;

                } else {

                    cost[j][k] = cost[j - 1][k - 1] - (color[h - j][i] == color[h - k][i + 1] ? 1 : 0);

                }

            }

        }

        Fill(dp,INF);

        dp[0][0] = 0;

        for (int j = 0; j <= h; ++j) {

            for (int k = 0; k <= h; ++k) {

                LL tmpA = j > 0 ? dp[j - 1][k] + cost[j - 1][k] : INF;

                LL tmpB = k > 0 ? dp[j][k - 1] + cost[j][k - 1] : INF;

                chmin(dp[j][k], min(tmpA, tmpB));

            }

        }

        sum += dp[h][h];

    }

    cout << sum << endl;

}

// Fastest known solution for this problem: