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)

#include <bits/stdc++.h>
#include <omp.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;


#define int ll
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());


typedef pair<int, int> pii;
typedef pair<string, string> pst;
typedef pair<pii, pii> piii;
typedef vector<piii> vpii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<pii> vpi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;
typedef pair<pld, pld> pldd;
typedef vector<pld> vpd;
typedef complex<double> base;


#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define my_abs(a) (((a) >= 0) ? (a) : -(a))
#define mp(a, b) make_pair(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define MODIK(a) (a >= M ? a - M : a)
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cout << (#x) << " = " << x << endl;
#ifdef ART_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif


const double PI = 2 * acos(0.0);

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";

istream &operator >> (istream &in, pii &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pii &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


istream &operator >> (istream &in, pld &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pld &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) <<
               ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void print(int a, int b, int n, vi h)
{
    ofstream cout("output.txt");
    cout << SZ(h) << ' ' << a << ' ' << b << '\n';
    for (int i = 0; i < SZ(h); ++i) {
        cout << h[i]  << ' ';
    }
    cout.close();
    exit(0);
}


inline bool break_time(ld time)
{
    return clock() > CLOCKS_PER_SEC * time;
}


ld get_rand_ld(ld l, ld r)
{
    uniform_real_distribution<double> gen(l, r);
    return gen(rng);
}


bool may(int E, ld T)
{
    if (E < 0)
        return 1;

    double eps = get_rand_ld(0, 1);
    return eps < exp(-E / T);
}


void smain();

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

#ifdef ART_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(2) << fixed;
    smain();
#ifdef ART_HOME
    cout << "\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
}


const int N = 100 + 10, B = 50, oo = 1e18 + 100, M = 998244353, LOG = 25;
const int M1 = 1000000093, x1 = 27162;
const int M2 = 1000000087, x2 = 241;
const ld pi = atan2(1, 0) * 2;


vst flex;
int used[N][N][2];


void smain()
{
    int n, m;
    cin >> n >> m;
    flex.resize(n);
    for (int i = 0; i < n; ++i)
        cin >> flex[i];

    deque<pii> q;
    q.push_back(mp(0, 0));
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            used[i][j][0] = used[i][j][1] = oo;

    used[0][0][0] =  used[0][0][1] = 1;

    while (SZ(q))
    {
        pii cur = q.front();
        q.pop_front();
        int v = min(used[cur.first][cur.second][0], used[cur.first][cur.second][1]);

        if (cur.first + 1 < n) {
            if (flex[cur.first + 1][cur.second] == flex[cur.first][cur.second] ||
                flex[cur.first][cur.second] != '.') {
                if (used[cur.first + 1][cur.second][0] == oo) {
                    q.push_front(mp(cur.first + 1, cur.second));
                    used[cur.first + 1][cur.second][0] = v;
                }
            } else {
                if (used[cur.first + 1][cur.second][1] == oo) {
                    q.push_back(mp(cur.first + 1, cur.second));
                    used[cur.first + 1][cur.second][1] = v + 1;
                }
            }
        }


        if (cur.second + 1 < m) {
            if (flex[cur.first][cur.second + 1] == flex[cur.first][cur.second] ||
                flex[cur.first][cur.second] != '.') {
                if (used[cur.first][cur.second + 1][0] == oo) {
                    q.push_front(mp(cur.first, cur.second + 1));
                    used[cur.first][cur.second + 1][0] = v;
                }
            } else {
                if (used[cur.first][cur.second + 1][1] == oo) {
                    q.push_back(mp(cur.first, cur.second + 1));
                    used[cur.first][cur.second + 1][1] = v + 1;
                }
            }
        }
    }

    int v = oo;
    if (used[n - 1][m - 1][0])
        v = used[n - 1][m - 1][0];

    if (used[n - 1][m - 1][1])
        v = min(v, used[n - 1][m - 1][1]);

//    int v = min(used[n - 1][m - 1], used[n - 1][m - 1][1]);
    if (flex[0][0] == '.')
        v--;

    cout << v;
}