Problem:
Score : 400 points

Problem Statement

Consider a grid with H rows and W columns of squares. Let (r, c) denote the square at the r-th row from the top and the c-th column from the left.
Each square is painted black or white.

The grid is said to be good if and only if the following condition is satisfied:

From (1, 1), we can reach (H, W) by moving one square right or down repeatedly, while always being on a white square.

Note that (1, 1) and (H, W) must be white if the grid is good.

Your task is to make the grid good by repeating the operation below. Find the minimum number of operations needed to complete the task. It can be proved that you can always complete the task in a finite number of operations.

Choose four integers r_0, c_0, r_1, c_1(1 \leq r_0 \leq r_1 \leq H, 1 \leq c_0 \leq c_1 \leq W). For each pair r, c (r_0 \leq r \leq r_1, c_0 \leq c \leq c_1), invert the color of (r, c) - that is, from white to black and vice versa.

Constraints

2 \leq H, W \leq 100

Input

Input is given from Standard Input in the following format:

H W
s_{11} s_{12} \cdots s_{1W}
s_{21} s_{22} \cdots s_{2W}
\vdots
s_{H1} s_{H2} \cdots s_{HW}

Here s_{rc} represents the color of (r, c) - # stands for black, and . stands for white.

Output

Print the minimum number of operations needed.

Sample Input 1

3 3
.##
.#.
##.

Sample Output 1

1

Do the operation with (r_0, c_0, r_1, c_1) = (2, 2, 2, 2) to change just the color of (2, 2), and we are done.

Sample Input 2

2 2
#.
.#

Sample Output 2

2

Sample Input 3

4 4
..##
#...
###.
###.

Sample Output 3

0

No operation may be needed.

Sample Input 4

5 5
.#.#.
#.#.#
.#.#.
#.#.#
.#.#.

Sample Output 4

4

Solution in JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

var input = require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");
var obj = {
  "list" : input, "index" : 0, "max" : input.length,
  "hasNext" : function(){return (this.index < this.max);},
  "next" : function(){
    if(!this.hasNext()){throw "NoSuchElementException:次に要素は無いよ";}
    var returnObj = this.list[this.index];
    this.index++;
    return returnObj;
  }
}
Main();
function next(){return obj.next();}
function nextInt(){return myconv(next(),1);}
function nextStrArray(){return myconv(next(),2);}//半角スペース分割
function nextIntArray(){return myconv(next(),4);}//半角スペース + 数値化
function nextCharArray(){return myconv(next(),6);}//1文字分割
function hasNext(){return obj.hasNext();}
function myout(t){console.log(t);}//標準出力
function myerr(t){console.error(t);}//標準エラー出力
//[no]要素の扱い。数値型
//不明値、異常時:引数そのまま返す　1:数値へ変換
//2:半角SPで分割 4:半角SPで分割し、数値配列へ
//6:1文字で分割  7:1文字で分割し、数値配列へ
//8:半角SPで結合 9:改行で結合 0:文字なしで結合
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map((a)=>Number(a));case 6:return i.split("");case 7:return i.split("").map((a)=>Number(a));case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}
function Main(){
  var one = nextIntArray();
  var H = one[0];
  var W = one[1];
  var list = new Array(H);
  var accessed = new Array(H);
  var counter = new Array(H);
  for(var i = 0; i < H; i++){
    list[i] = nextCharArray();
    accessed[i] = new Array(W).fill(false);
    counter[i] = new Array(W).fill(0);
  }
  //最小で何個#にひっかかるか
  var output = 10007;
  var dy = [-1,0,1,0];
  var dx = [0,-1,0,1];
  var queue = [[0,0,0]];
  accessed[0][0] = true;
  while(queue.length > 0){
    var tmp = queue.shift();
    var y = tmp[0];
    var x = tmp[1];
    var c = tmp[2];
    accessed[y][x] = true;
    if(list[y][x] == "#"){
      c++;
      counter[y][x] = Math.max(counter[y][x],c);
    }
    if(y == H - 1 && x == W - 1){
      output = Math.min(output,c);
    }
    for(var i = 0; i < 4; i++){
      var nextY = y + dy[i];
      var nextX = x + dx[i];
      if(nextY >= 0 && nextY < H && nextX >= 0 && nextX < W){
        if(!accessed[nextY][nextX]){
          
          queue.push([nextY,nextX,c]);
        }
      }
    }
  }
  myout(output);
  //myerr(counter);
}