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: Wrong Answer)

  /// DONOT COPY PASTA WITHOUT AUTHOR'S PERMISSION ///
 ///                   S S S                      ///
///                1 7 0 7 0 5 3                 ///

#include<bits/stdc++.h>
#include<string>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <stdlib.h>
#include<string>
///STL
#define mp make_pair
#define ss second
#define ff first
#define pb push_back
#define ist insert
///ONLINE JUDGE
#define io ios_base::sync_with_stdio;cin.tie(0);cout.tie(0)
#define READ freopen("input.txt","r", stdin)
#define case(i) "Case "<<i++<<": "
#define nscase(i) "Case "<<i++<<":"
///c shortcuts
#define sns(s) scanf("%s",&s)
#define snc(c) scanf("%c",&c)
#define snd(a) scanf("%d",&a)
#define snl(a) scanf("%lld",&a)
#define snul(a) scanf("%ull",&a)
///Loops
#define mem(a,b) memset(a, b, sizeof(a))
#define rep(i,k,n) for(int i=k;i<n;i++)
#define rev(j,n,k) for(int j=n;j>=k;j--)
#define TC int tc;cin>>tc;while(tc--)

///GCD LCM
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a / __gcd(a, b) ) * b)

///CONSTANTS
#define PI 2*acos(0.0)
#define MAX 1000000000
#define inf 1e18

///functions
#define base(n,b) floor(log(n)/log(b))+1
#define onecnt(x) __builtin_popcount(x)
#define to_binary(n,b) bitset<b>(n).to_string()

#define endl '\n'
#define esc cout<<"\n"

/*------------------------------Graph Moves----------------------------*/
///const int fx[]= {+1,-1,+0,+0};
///const int fy[]= {+0,+0,+1,-1};
///const int fx[]= {+0,+0,+1,-1,-1,+1,-1,+1};  // Kings Move
///const int fy[]= {-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
///const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
///const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long,long> pll;
typedef std::map<long long,int> mli;
typedef std::map<long long,long long> mll;
typedef std::map<int,int> mii;

///PBDS
using namespace __gnu_pbds;
typedef tree<
ll,
null_type,//set to another type while using map
less<ll>,
rb_tree_tag,
tree_order_statistics_node_update> ordered_set;

template <typename T>
std::string toString( T Number )
{
    std::ostringstream ss;
    ss << Number;
    return ss.str();
}

///constants
const ll ten5=100000;
const ll ten6 =1000000;
const ll ten7 =10000000;
const ll ten8 =100000000;
const ll ten9 =1000000000;
const ll mod=ten9+7;


int mat[105][105];
int main()
{
    io;

    int n,m;
    cin>>n>>m;

    string s[n];
    for(int i=0;i<n;i++){
        cin>>s[i];
    }

    if(s[0][0]=='#') mat[0][0]=1;
    for(int i=1;i<m;i++){
        int x=0;
        if(s[0][i]=='#')x++;
        mat[0][i]=mat[0][i-1]+x;
    }
    for(int i=1;i<n;i++){
        int x=0;
        if(s[i][0]=='#')x++;
        mat[i][0]=mat[i-1][0]+x;
    }
    for(int i=1;i<n;i++){
        for(int j=1;j<m;j++){
            int l,r,x;


            l=mat[i][j-1];


            r=mat[i-1][j];
            if(s[i][j]=='#') x=1;
            else x=0;

            mat[i][j]=min(l,r)+x;
            //cout<<mat[i][j]<<" ";
        }
        //esc;
    }

    cout<<mat[n-1][m-1];
}