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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int MIN(int a,int b){return a==0?b:(a<b?a:b);}
char s[100010][3];
int h,d[310][310][310]={1};
int f(int x,int a,int b,int c){
  int co=0,i;
  if(x==0){
    for(i=0;i<h-a;i++){
      if(a+i>=b&&s[i][0]==s[i+a-b][1])co++;
    }
  }
  if(x==1){
    for(i=0;i<h-b;i++){
      if(b+i>=a&&s[i][1]==s[i+b-a][0])co++;
      if(b+i>=c&&s[i][1]==s[i+b-c][2])co++;
    }
  }
  if(x==2){
    for(i=0;i<h-c;i++){
      if(c+i>=b&&s[i][2]==s[i+c-b][1])co++;
    }
  }
  //printf("%d %d %d %d %d\n",x,a,b,c,co);
  return co;
}
int main(){
  int w,i,j,k;
  scanf("%d %d",&h,&w);
  if(w>3){
    printf("0\n");
    return 0;
  }
  for(i=0;i<h;i++)scanf("%s",s[i]);
  for(i=0;i<=h;i++){
    for(j=0;j<=h;j++){
      for(k=0;k<=h;k++){
	d[i+1][j][k]=MIN(d[i+1][j][k],d[i][j][k]+f(0,i,j,k));
	d[i][j+1][k]=MIN(d[i][j+1][k],d[i][j][k]+f(1,i,j,k));
	d[i][j][k+1]=MIN(d[i][j][k+1],d[i][j][k]+f(2,i,j,k));
      }
    }
  }
  for(i=0;i<=h;i++){
    for(j=0;j<=h;j++){
      //for(k=0;k<=h;k++)printf("%d %d %d %d\n",i,j,k,d[i][j][k]);
    }
  }
  printf("%d\n",d[h][h][h]-1);
  return 0;
}