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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define ll long long

const int N = 310;

bitset<300> mask[N][26];
int dp[N][N][N];

int go(int i, int j, int k, int h) {
	if (i == h && j == h && k == h) return 0;
	if (dp[i][j][k] != -1) return dp[i][j][k];

	int ans = INT_MAX;
	if (i < h) {
		int cost = 0;
		for (int c = 0; c < 26; c++) {
			cost += ((mask[0][c] >> i) & (mask[1][c] >> j)).count();
		}
		ans = min(ans, cost + go(i + 1, j, k, h));
	}

	if (j < h) {
		int cost = 0;
		for (int c = 0; c < 26; c++) {
			cost += ((mask[0][c] >> i) & (mask[1][c] >> j)).count();
			cost += ((mask[1][c] >> j) & (mask[2][c] >> k)).count();
		}
		ans = min(ans, cost + go(i, j + 1, k, h));
	}

	if (k < h) {
		int cost = 0;
		for (int c = 0; c < 26; c++) {
			cost += ((mask[1][c] >> j) & (mask[2][c] >> k)).count();
		}
		ans = min(ans, cost + go(i, j, k + 1, h));
	}

	return dp[i][j][k] = ans;
}

int main() {
	memset(dp, -1, sizeof dp);

	int h, w;
	scanf("%d %d", &h, &w);

	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			char c;
			scanf(" %c", &c);
			mask[j][c - 'a'].set(h - i - 1, true);
		}
	}

	printf("%d\n", go(0, 0, 0, h));

	return 0;
}