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: Accepted)

//DUEL
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define x first
#define y second
#define pii pair<int,int>
#define pb push_back
#define eb emplace_back
#pragma GCC optimize("unroll-loops")
#define shandom_ruffle(a, b) shuffle(a, b, rng)
#define vi vector<int>
#define vl vector<ll>
#define popcnt __builtin_popcount
#define popcntll __builtin_popcountll
#define all(a) begin(a),end(a)

using namespace std;
using namespace __gnu_pbds;

using ll=long long;
using ull=unsigned long long;
using ld=long double;
int MOD=1000000007;
int MOD2=998244353;
vector<int> bases;
const ll LLINF=1ll<<60;
const char en='\n';

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void yes() {cout<<"YES"<<en; exit(0);}
void no() {cout<<"NO"<<en; exit(0);}
inline int rund() {int x576363482791fuweh=rng();return abs(x576363482791fuweh)%RAND_MAX;}
template<class T>
void prVec(vector<T> w,bool fl=false)
{
	cout<<w.size()<<en;
	for (int i=0;i<int(w.size())-1;++i) cout<<w[i]<<' ';
	if (w.size()) cout<<w[w.size()-1]<<en;
	if (fl) cout<<flush;
}

void M998()
{
	swap(MOD,MOD2);
}

ll raand()
{
	ll a=rund();
	a*=RAND_MAX;
	a+=rund();
    return a;
}

#define rand raand

ll raaand()
{
	return raand()*(MOD-7)+raand();
}

string to_upper(string a)
{
	for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A';
	return a;
}

string to_lower(string a)
{
	for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A';
	return a;
}

ll sti(string a,int base=10)
{
	ll k=0;
	for (int i=0;i<(int)a.size();++i)
	{
		k*=base;
		k+=a[i]-'0';
	}
	return k;
}

template<class T>
void eras(vector<T>& a,T b)
{
	a.erase(find(a.begin(),a.end(),b));
}

string its(ll k,int base=10)
{
	if (k==0) return "0";
	string a;
	while (k)
	{
		a.push_back((k%base)+'0');
		k/=base;
	}
	reverse(a.begin(),a.end());
	return a;
}

ll min(ll a,int b)
{
	if (a<b) return a;
	return b;
}

ll min(int a,ll b)
{
	if (a<b) return a;
	return b;
}

ll max(ll a,int b)
{
	if (a>b) return a;
	return b;
}

ll max(int a,ll b)
{
	if (a>b) return a;
	return b;
}

ll gcd(ll a,ll b)
{
	if (b==0) return a;
	return gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
	return a/gcd(a,b)*b;
}

template<class T,class K>
pair<T,K> mp(T a,K b)
{
	return make_pair(a,b);
}

inline int mult(ll a,ll b)
{
	return (a*b)%MOD;
}

inline int pot(int n,int k)
{
	if (k==0) return 1;
	ll a=pot(n,k/2);
	a=mult(a,a);
	if (k%2) return mult(a,n);
	else return a;
}

int divide(int a,int b)
{
	return mult(a,pot(b,MOD-2));
}

inline int sub(int a,int b)
{
	if (a-b>=0) return a-b;
	return a-b+MOD;
}

inline int add(int a,int b)
{
	if (a+b>=MOD) return a+b-MOD;
	return a+b;
}

bool prime(ll a)
{
	if (a==1) return 0;
	for (int i=2;i<=round(sqrt(a));++i)
	{
		if (a%i==0) return 0;
	}
	return 1;
}

const int N=310;
int n,m,dp[N][N],r[N][N];
string h[N];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	for (int i=0;i<10;++i) bases.push_back(rand()%(MOD-13893829*2)+13893829);
	cin>>n>>m;
	++m;
	for (int i=0;i<n;++i)
	{
		cin>>h[i];
		if (i<n-1) h[i].pb('/');
		else h[i].pb('.');
	}
	memset(r,63,sizeof(r));
	r[n-1][m-1]=0;
	for (int i=n-1;i>=0;--i)
	{
		for (int j=m-2;j>=0;--j)
		{
			for (int k=i;k<n;++k) for (int l=j;l<m;++l) dp[k][l]=0;
			dp[i][j]=1;
			for (int k=i;k<n;++k) for (int l=j;l<m-1;++l) if (h[k][l]==h[i][j] && dp[k][l])
			{
				dp[k+1][l]=1;
				dp[k][l+1]=1;
			}
			for (int k=i;k<n;++k) for (int l=j;l<m;++l) if (dp[k][l])
			{
				//cout<<i<<' '<<j<<' '<<k<<' '<<l<<' ';
				if (h[i][j]=='#') r[i][j]=min(r[i][j],r[k][l]+1);
				else r[i][j]=min(r[i][j],r[k][l]);
				//cout<<r[i][j]<<en;
			}
		}
	}
	//for (int i=0;i<n;++i,cout<<en) for (int j=0;j<m-1;++j) cout<<r[i][j];
	cout<<r[0][0];
}