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>
const long long INF = LLONG_MAX / 2;
const long long MOD = 1000000007;
#define FOR(i, r, n) for(int i=(ll)(r); i<(ll)(n); i++) 
#define REP(i, n) FOR(i, (0), n)
#define ll long long int
using namespace std;
ll ans = 0, sum = 0, n, k, counter = 0;
string s;
vector<ll> v;
vector<pair<ll, ll>> vp;
vector<vector<ll>> vv(60, vector<ll>(60, INF));

vector<vector<ll>> vc(300, vector<ll>(300, 0));
vector<ll> cost(3,0);

int main()
{
	ll h, w;
	cin >> h >> w;
	if(w!=3)
	{
		cout << "-1" << endl;
		return 0;
	}
	REP(i, h) {
		cin >> s;
		REP(j, w)
			vc[i][j]=s[j];
	}
	REP(i,h-1)
	{
		FOR(j, i, h)
			if (vc[i][0] == vc[j][1]) cost[0]++;
	}
	REP(i, h-1)
	{
		FOR(j, i, h) {
			if (vc[i][1] == vc[j][0]) cost[1]++;
			if (vc[i][1] == vc[j][2]) cost[1]++;
		}
	}
	REP(i, h-1)
	{
		FOR(j, i, h)
			if (vc[i][2] == vc[j][1]) cost[2]++;
	}
	sort(cost.begin(), cost.end());
	cout << cost[0] << endl;
}