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 <bits/stdc++.h>
using namespace std;

#define MAX 302
int h;
int w;

char buf[MAX][MAX];


string x;
string y;
int dp[302][302];
int same[302][302];
int calc(){
	for (int i = 0; i <= x.size(); i++){
		for (int j = 0; j <= y.size(); j++){
			dp[i][j] = INT_MAX;
			same[i][j] = 0;
		}
	}
	for (int i = x.size() - 1; i >= 0; i--){
		for (int j = y.size() - 1; j >= 0; j--){
			same[i][j] = (x[i] == y[j]);
			same[i][j] += same[i + 1][j + 1];
		}
	}
	dp[0][0] = 0;
	for (int i = 0; i <= x.size(); i++){
		for (int j = 0; j <= y.size(); j++){
			if (i < x.size()){
				dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + same[i][j]);
			}
			if (j < y.size()){
				dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + same[i][j]);
			}
		}
	}
	return dp[x.size()][y.size()];
}
int main(){
	cin >> h >> w;
	for (int i = 0; i < h; i++){
		scanf("%s", buf[i]);
	}
	int sum = 0;
	for (int i = 1; i < w; i++){
		x.clear();
		y.clear();
		for (int j = 0; j < h; j++){
			x.push_back(buf[j][i-1]);
			y.push_back(buf[j][i]);
		}
		reverse(x.begin(), x.end());
		reverse(y.begin(), y.end());
		sum += calc();
	}
	cout << sum << endl;
	return 0;
}