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: Time Limit Exceeded)

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <map>
#include <numeric>
#include <cmath>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <complex>
#include <string.h>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <iomanip>
#include <sys/time.h>
#include <random>
using namespace std;

#define endl '\n'
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define UNIQ(v) (v).erase(unique((v).begin(), (v).end()), (v).end())

typedef long long ll;
typedef long double ld;
typedef pair<int, int> P;
typedef complex<double> comp;
typedef vector< vector<ld> > matrix;
struct pairhash {
public:
    template<typename T, typename U>
    size_t operator()(const pair<T, U> &x) const {
	size_t seed = hash<T>()(x.first);
	return hash<U>()(x.second) + 0x9e3779b9 + (seed<<6) + (seed>>2);
    }
};
const int inf = 1e9 + 9;
const ll mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);

int h, w;
char c[310][310];

int dp[310][310];
int idx = 0;

int calc(int h0, int h1) {
    if (h0 == 0 || h1 == 0) return 0;
    if (dp[h0][h1] < 0) {                
        int e = 0;
        int s0, s1;
        if (h0 < h1) {
            s0 = 0;
            s1 = h1 - h0;
        } else {
            s0 = h0 - h1;
            s1 = 0;
        }
        for (int i = 0; i < min(h0, h1); i++, s0++, s1++) {
            if (c[s0][idx] == c[s1][idx+1]) e++;
        }
        dp[h0][h1] = min(calc(h0-1, h1), calc(h0, h1-1)) + e;
    }
    return dp[h0][h1];
}

int solve() {
    int res = 0;
    for (int i = 0; i < w-1; i++) {
        idx = i;
        memset(dp, -1, sizeof(dp));
        res += calc(h, h);
    }
    return res;
}

void input() {
    cin >> h >> w;
    string s;
    for (int i = 0; i < h; i++) {
        cin >> s;
        for (int j = 0; j < w; j++)
            c[i][j] = s[j];
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(15);

    input();
    cout << solve() << endl;
}