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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

int d[301][301][301];

int onetwo[300][300];
int twothr[300][300];
int onethr[300][300];

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(NULL);

    int h, w;
    cin >> h >> w;

    if (w == 3) {
        int n = h;
        string a(n, '0'), b(n, '0'), c(n, '0');

        for (int i = 0; i < n; i++) {
            cin >> a[i] >> b[i] >> c[i];
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int ii = i, jj = j;

                for (int k = min(i, j); k >= 0; k--) {
                    onetwo[i][j] += (a[ii] == b[jj]);
                    ii--;
                    jj--;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int ii = i, jj = j;

                for (int k = min(i, j); k >= 0; k--) {
                    twothr[i][j] += (b[ii] == c[jj]);
                    ii--;
                    jj--;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int ii = i, jj = j;

                for (int k = min(i, j); k >= 0; k--) {
                    onethr[i][j] += (a[ii] == c[jj]);
                    ii--;
                    jj--;
                }
            }
        }

        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
                for (int k = 0; k <= n; k++) {
                    if (i == 0 && j == 0 && k == 0) {
                        continue;
                    }
                    d[i][j][k] = 1000 * 1000;

                    if (i > 0) {
                        d[i][j][k] = min(d[i][j][k], d[i - 1][j][k] + (j == 0 ? 0 : onetwo[i - 1][j - 1]));
                    }
                    if (k > 0) {
                        d[i][j][k] = min(d[i][j][k], d[i][j][k - 1] + (j == 0 ? 0 : twothr[j - 1][k - 1]));
                    }
                    if (j > 0) {
                        d[i][j][k] = min(d[i][j][k], d[i][j - 1][k] + (i == 0 ? 0 : onetwo[i - 1][j - 1]) + (k == 0 ? 0 : twothr[j - 1][k - 1]));
                    }
                }
            }
        }
        cout << d[n][n][n] << '\n';
    } else {
        throw;
    }

    return 0;
}