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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<time.h>
#include<random>
#include<array>
using namespace std;
using ll = long long;
#define rep(i, a, b) for(int i = a; i < b; i++)
#define rrep(i, a, b) for(int i = b - 1; i >= a; i--)
#define ALL(a) a.begin(), a.end()
using pii = pair<int,int>;
using piii = pair<pii,int>;
using pll = pair<long long, long long>;
using plll = pair<pll, long long>;
// #pragma GCC optimize("Ofast")
#define pcnt __builtin_popcount
#define buli(x) __builtin_popcountll(x)
#define pb push_back
#define mp make_pair
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define isSquare(x) (sqrt(x)*sqrt(x) == x)
template<class T>bool chmax(T &a, const T &b) {if(a<b){a = b; return 1;} return 0; };
template<class T>bool chmin(T &a, const T &b) {if(a>b){a = b; return 1;} return 0; };
inline void in(void){return;}
template <typename First, typename... Rest> void in(First& first, Rest&... rest){cin >> first;in(rest...);return;}
inline void out(void){cout << "\n";return;}
template <typename First, typename... Rest> void out(First first, Rest... rest){cout << first << " ";out(rest...);return;}
const double EPS = 1e-9;
const int mod = 1e9 + 7;
const int INF = 1e9;
const long long INFLL = 1e18;
void iosetup() {
    cin.tie(nullptr);ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    cerr << fixed << setprecision(10);
}
template< typename T1, typename T2 >
ostream &operator<<(ostream &os, const pair< T1, T2 >& p) {
    os << p.first << " " << p.second;
    return os;
}
template< typename T1, typename T2 >
istream &operator>>(istream &is, pair< T1, T2 > &p) {
    is >> p.first >> p.second;
    return is;
}
template< typename T >
ostream &operator<<(ostream &os, const vector< T > &v) {
    for(int i = 0; i < (int) v.size(); i++) {
        os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
}
template< typename T >
istream &operator>>(istream &is, vector< T > &v) {
    for(T &in : v) is >> in;
    return is;
}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S, T>& t){return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S, T>& t){return pair<S,T>(s.first-t.first, s.second-t.second);}
template<class S, class T> pair<S,T> operator*(const pair<S,T> &s, const S& t){return pair<S,T>(s.first*t, s.second*t);}
template <typename T> void Exit(T first){cout << first << endl;exit(0); };
template< int mod > struct ModInt {
    unsigned x; ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {if((x += p.x) >= mod) x -= mod;return *this;}
    ModInt &operator-=(const ModInt &p) {if((x += mod - p.x) >= mod) x -= mod;return *this;}
    ModInt &operator*=(const ModInt &p) {x = (int) (1LL * x * p.x % mod);return *this;}
    ModInt &operator/=(const ModInt &p) {*this *= p.inverse();return *this;}
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    ModInt inverse() const {int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) { t = a / b; swap(a -= t * b, b); swap(u -= t * v, v); }return ModInt(u);}
    ModInt pow(int64_t n) const {ModInt ret(1), mul(x); while(n > 0) {if(n & 1) ret *= mul;mul *= mul;n >>= 1;}return ret;}
    friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x;}
    friend istream &operator>>(istream &is, ModInt &a) { int64_t t; is >> t; a = ModInt< mod >(t); return (is); }
    static int get_mod() { return mod; }
}; using modint = ModInt< mod >;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const pii dxy[4] = {pii(1,0), pii(0, 1), pii(-1, 0), pii(0, -1)};
const int calender[13] = {0, 31,28,31,30,31,30,31,31,30,31,30,31};
const int ucalender[13] = {0, 31,29,31,30,31,30,31,31,30,31,30,31};
bool range(int a, int b, int x){if(a <= x and x < b)return true;else return false;}
bool range(int a, int b, int c, int d, pii p){if(a <= p.first and p.first < b and c <= p.second and p.second < d) return true;else return false;}

// int cost[310][800][310];
int dp[310][310];
int main(){
    iosetup();
    int H, W; cin >> H >> W;
    vector<string> C(H); cin >> C;
    ll ans = 0;
    rep(i, 0, W - 1){
        string S = "", T = "";
        rep(j, 0, H) S += C[j][i];
        rep(j, 0, H) T += C[j][i+1];
        vector<vector<int> > cost(H+1, vector<int>(H+1, 0));
        // cost[i][j] := 左がi個，みぎがj個のコスト
        rep(j, 1, H + 1){
            rep(k, 1, H + 1){
                cost[j][k] = cost[j-1][k-1] + (S[j-1] == T[k-1]);
            }
        }
        // rep(j, 0, H + 1) cerr << cost[j] << endl;

        rep(j, 0, H + 1) rep(k, 0, H + 1) dp[j][k] = INF;
        dp[0][0] = 0;
        rep(j, 0, H + 1){
            rep(k, 0, H + 1){
                chmin(dp[j+1][k], dp[j][k] + cost[H-j][H-k]);
                chmin(dp[j][k+1], dp[j][k] + cost[H-j][H-k]);
            }
        }
        ans += dp[H][H];
    }
    cout << ans << endl;


    return 0;
}