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>
using namespace std;

#define CPP_STR(x) CPP_STR_I(x)
#define CPP_CAT(x,y) CPP_CAT_I(x,y)
#define CPP_STR_I(args...) #args
#define CPP_CAT_I(x,y) x ## y

using i8  = int8_t;
using u8  = uint8_t;
using i16 = int16_t;
using u16 = uint16_t;
using i32 = int32_t;
using u32 = uint32_t;
using i64 = int64_t;
using u64 = uint64_t;

using f32 = float;
using f64 = double;

#define FOR(i, start, end) for(i64 i = (start), CPP_CAT(i,xxxx_end)=(end); i < CPP_CAT(i,xxxx_end); ++i)
#define REP(i, n) FOR(i, 0, n)
#define REPS(i, n) for(i64 i = 1; i <= n; ++i)
#define RREP(i, n) for(i64 i = n - 1; i >= 0 ; --i)
#define RREPS(i, n) for(i64 i = n; i >= 1; --i)

#define ALL(f,c,...) (([&](decltype((c)) cccc) { return (f)(std::begin(cccc), std::end(cccc), ## __VA_ARGS__); })(c))

#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

#define pb push_back
#define mp make_pair

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

template<typename C>
i64 SIZE(const C& c) { return static_cast<i64>(c.size()); }

template<typename T, size_t N>
i64 SIZE(const T (&)[N]) { return static_cast<i64>(N); }


struct ProconInit {
    static constexpr int IOS_PREC = 15;
    static constexpr bool AUTOFLUSH = false;

    ProconInit() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(IOS_PREC);
        if(AUTOFLUSH)
            cout << unitbuf;
    }
} PROCON_INIT;

#define MAX_H 100
#define MAX_W 100
#define INF 10'000'000
i64 H, W;
i64 G[MAX_H][MAX_W];
i64 memo[MAX_H][MAX_W][2];


i64 dfs(i64 y, i64 x, i64 prev) {
    if (y == H - 1 && x == W - 1) {
        if (prev == 0 && G[y][x] == 1) return 1;
        else return 0;
    }

    if (memo[y][x][prev] >= 0) {
        return memo[y][x][prev];
    }

    i64 minc = INF;
    if (prev == 0 && G[y][x] == 1) {
        if (y + 1 < H) chmin(minc, dfs(y + 1, x, G[y][x]) + 1);
        if (x + 1 < W) chmin(minc, dfs(y, x + 1, G[y][x]) + 1);
    } else {
        if (y + 1 < H) chmin(minc, dfs(y + 1, x, G[y][x]));
        if (x + 1 < W) chmin(minc, dfs(y, x + 1, G[y][x]));
    }
    return memo[y][x][prev] = minc;
}

int main() {
    cin >> H >> W;
    REP(i, H) REP(j, W) {
        char c; cin >> c;
        G[i][j] = (c == '#');
    }

    REP(i, H) REP(j, W) REP(k, 2) memo[i][j][k] = -1;

    i64 minc = dfs(0, 0, 0);
    cout << minc << endl;
    return 0;
}