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 Rust (1.15.1) (judge verdict: Accepted)

macro_rules !read_value {($iter :expr ,($($t :tt ) ,*) ) =>{($(read_value !($iter ,$t ) ) ,*) } ;($iter :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($iter ,$t ) ) .collect ::<Vec <_ >>() } ;($iter :expr ,{chars :$base :expr } ) =>{read_value !($iter ,String ) .chars () .map (|c |(c as u8 -$base as u8 ) as usize ) .collect ::<Vec <usize >>() } ;($iter :expr ,{char :$base :expr } ) =>{read_value !($iter ,{chars :$base } ) [0 ] } ;($iter :expr ,chars ) =>{read_value !($iter ,String ) .chars () .collect ::<Vec <char >>() } ;($iter :expr ,char ) =>{read_value !($iter ,chars ) [0 ] } ;($iter :expr ,usize1 ) =>{read_value !($iter ,usize ) -1 } ;($iter :expr ,$t :ty ) =>{$iter .next () .unwrap () .parse ::<$t >() .unwrap () } ;}
macro_rules !input_inner {($iter :expr ) =>{} ;($iter :expr ,) =>{} ;($iter :expr ,mut $var :ident :$t :tt $($r :tt ) *) =>{let mut $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,mut $var :ident $($r :tt ) *) =>{input_inner !{$iter ,mut $var :usize $($r ) *} } ;($iter :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,$var :ident $($r :tt ) *) =>{input_inner !{$iter ,$var :usize $($r ) *} } ;}
macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;(iter =$iter :ident ,$($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut $iter =s .split_whitespace () ;input_inner !{$iter ,$($r ) *} } ;($($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut iter =s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;}
macro_rules! chmin {
    ($l:expr, $r:expr) => {{
        let x = $r;
        $l = std::cmp::min($l, x);
    }};
}
fn main() {
    input!(h, w, c: [chars; h]);
    let mut ans = 0;
    for i in 0..w - 1 {
        let ni = i + 1;
        let mut dp = vec![vec![std::usize::MAX; h + 1]; h + 1];
        dp[0][0] = 0;
        let mut cost = vec![vec![0; h + 1]; h + 1];
        for j in 0..h {
            cost[j][0] = (0..h - j).filter(|&k| c[k][i] == c[k + j][ni]).count();
        }
        for j in 1..h {
            cost[0][j] = (0..h - j).filter(|&k| c[k + j][i] == c[k][ni]).count();
        }
        for j in 0..h {
            for k in 0..h {
                let nj = j + 1;
                let nk = k + 1;
                cost[nj][nk] = cost[j][k] - (c[h - nj][i] == c[h - nk][ni]) as usize;
            }
        }
        for j in 0..h + 1 {
            for k in 0..h + 1 {
                let nj = j + 1;
                let nk = k + 1;
                if j < h {
                    chmin!(dp[nj][k], dp[j][k] + cost[j][k]);
                }
                if k < h {
                    chmin!(dp[j][nk], dp[j][k] + cost[j][k]);
                }
            }
        }
        ans += dp[h][h];
    }
    println!("{}", ans);
}