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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

//#define _GLIBCXX_DEBUG//TLEの原因になるので注意！！！！！！！！！！！
#include<cmath>
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
vector<int> arr;
stack<int> st;
queue<int> qu;
queue<pair<int,int> > qu2;
priority_queue<int> pq;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define rep2(i,n) for(int i=1; i<=(int)(n); i++)
#define mins(x,y) x=min(x,y)
#define maxs(x,y) x=max(x,y)
typedef set<int> set_t;
typedef set<string> set_g;
typedef complex<double> xy_t;
static const int NIL = -1;
static const int INF = 1000000007;
#define mp make_pair
#define sz(x) int(x.xise())
#define mod 1000000007
#define reps(i,s,n) for(int i = s; i < n; i++)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
deque<int> deq;
#define fi first
#define se second
//#define int long long

//cerrはコメントアウトしなくても通る//ただしTLEを引き起こすことがある

ll N,M,K,Q,A,B;
string S;
//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
//const ll INF = 1LL << 60;
//typedef pair<ll,ll> P;
typedef pair<int, int> P;
bool chmx(int &a, int b) {
	if (b > a) {
		a = b;
		return true;
	}
	return false;
}

bool chmn(int &a, int b) {
	if (b < a) {
		a = b;
		return true;
	}
	return false;
}

string s[105];
int t[105][105];
vector<int> to[10005];

int main(){
    int h,w;
    cin >> h>>w;
    rep(i,h){
        cin >> s[i];
    }
    rep(i,h){
        rep(j,w){
            if(s[i][j]=='#'){//black
                if((s[i+1][j]=='#')||(s[i][j+1]=='#')){
                    t[i][j]=3;//blackかつ未来がある
                }else{
                    t[i][j]=2;//ない
                }
            }else if(s[i][j]=='.'){
                if((s[i+1][j]=='.')||(s[i][j+1]=='.')){
                    t[i][j]=1;//whiteかつ未来がある
                }else{
                    t[i][j]=0;//ない
                }
            }
        }
    }
    /*
    rep(i,h){
        rep(j,w){
            cerr<<t[i][j]<<' ';
        }
        cerr<<endl;
    }
    */
    int z=h*w;
    int dis[z][z];//iからjにいく
    rep(i,z){
        rep(j,z){
            if(i==j){
                dis[i][j]=0;
            }else if(j-i==1&&(j%w!=0)){
                int H=i/w;
                int W=i%w;
                if(t[H][W]==1||t[H][W]==3){
                    if(s[H][W]==s[H][W+1]){
                        dis[i][j]=0;
                        to[i].push_back(j);
                    }
                }else{
                    dis[i][j]=1;
                    to[i].push_back(j);
                }
            }else if(j-i==w){
                int H=i/w;
                int W=i%w;
                if(t[H][W]==1||t[H][W]==3){
                    if(s[H][W]==s[H+1][W]){
                        dis[i][j]=0;
                        to[i].push_back(j);
                    }
                    
                }else{
                    dis[i][j]=1;
                    to[i].push_back(j);
                }
            }else{
                dis[i][j]=INF;
            }
        }
    }
    int d[z];
    rep(i,z){
        d[i]=INF;
    }
    d[0]=0;
    pq.push(0);
    while(!pq.empty()){
        int k=pq.top();
        pq.pop();
        for(int u=0;u<to[k].size();u++){
            int p=to[k][u];
            if(d[p]>d[k]+dis[k][p]){
                d[p]=d[k]+dis[k][p];
                pq.push(p);
            }
        }
    }

    int ans=d[z-1];
    ans=ans/2;
    if((t[0][0]==3||t[0][0]==2)&&(t[h-1][w-1]==3||t[h-1][w-1]==2)){
        ans++;
    }
    if(s[0][0]!=s[h][w]){
        ans++;
    }
    
    cout<<ans<<endl;
    /*
    rep(i,z){
        for(int j=0;j<to[i].size();j++){
            cerr<<to[i][j]<<' ';
        }
        cerr<<endl;
    }
    */
    /*
    rep(i,z){
        cerr<<d[i]<<endl;
    }
    */
    return 0;
}