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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, s, n) for (int i = s; i < n; i++)
#define rp(i,n) for (int i=0;i<n;i++)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define all_acc(x) (x).begin(), (x).end(), 0LL
#define int long long
#define absSort(v) sort(a.begin(), a.end(), [](int i, int j) -> bool { return abs(i) < abs(j); });
using namespace std;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
using ll = long long;
double pi=3.14159265359;

//出力するよ
void output_tate(vector<int> s){
    int n = s.size();
    for (int i = 0; i < n; i++){
        //if (i != 0)
        //    cout << " ";
        cout << s[i]<<endl;
    }
    //cout << endl;
}

void output_yoko(vector<int> s){
    int n = s.size();
    for (int i = 0; i < n; i++){
        if (i != 0)
            cout << " ";
        cout << s[i];//<<endl;
    }
    cout << endl;
}
int MOD = pow(10, 9) + 7;

//幅優先探索
//disは各座標への距離。行けない場合は-1にしている。
//chaは各座標の情報。文字で格納している。'.'はすすめる。'#'は壁(黒)。
//sはスタートの座標
//gはゴールの座標
vector<vector<int>> bfs(vector<vector<int>> dis,vector<vector<char>>cha,pair<int,int>s,pair<int,int>g){
    int h=dis.size();
    int w=dis.at(0).size();

    queue<pair<int,int>> Q;
    Q.push(s);
    dis.at(s.first).at(s.second)=0;
    if(cha[0][0]=='#') dis[0][0]=1;
    else dis[0][0]=0;

    while(!Q.empty()){
        int y,x;//今の座標
        y=Q.front().first; x=Q.front().second;
        Q.pop();

        //今回は後ろには戻らない。というか、0,0スタートなら戻る必要無さそう。
        int dy[4]={1,0,-1,0};
        int dx[4]={0,1,0,-1};
        for(int i=0;i<2;i++){
            //ey,exは変更後の座標。csは変える候補の値。
            int ey=y+dy[i];
            int ex=x+dx[i];
            int cs;
            //範囲外はスルー
            if(ex<0||ey<0||ex>=w||ey>=h) continue;
            //白から黒へ移るときにカウントが１増える
            if(cha[y][x]=='.' && cha[ey][ex]=='#') cs=dis[y][x]+1;
            else cs=dis[y][x];
            
            //未探索か、候補値の方が小さい場合、変更して変更先をキューに打ち込む
            if(dis[ey][ex]==-1 || cs<dis[ey][ex]){
                Q.push({ey,ex});
                dis[ey][ex]=cs;
            }
        }
    }

    return dis;
}

signed main() {
    int h,w; cin>>h>>w;
    vector<vector<char>> grid(h,vector<char>(w));
    vector<vector<int>> dis(h,vector<int>(w,-1));
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++) cin>>grid[i][j];
    }

    dis=bfs(dis,grid,make_pair(0,0),make_pair(h-1,w-1));

    cout<<dis[h-1][w-1]<<endl;
}