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;
using ll = long long;
const ll INF = ll(1e12);
const ll mod = ll(1e9+7);
const double pi = acos(-1);
#define rep0(i,n) for(ll (i) = 0; (i) < (n); ++(i))
#define rrep0(i,n) for(ll (i) = (n) - 1; (i) >= 0; --(i))
#define rep1(i,n) for(ll (i) = 1; (i) <= (n); ++(i))
#define rrep1(i,n) for(ll (i) = (n); (i) >= 1; --(i))
#define nfor(i,a,b) for(ll (i) = (a); (i) < (b); ++(i))
#define pf(x) cout << (x) << endl
#define all(x) (x).begin(),(x).end()
#define yes pf("Yes")
#define no pf("No")

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

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

ll factorial(ll n){
    ll ans = 1;
    rep1(i, n){
        ans *= i;
        ans %= mod;
    }
    return ans;
}

ll power(ll a, ll b){
    ll ans = 1;
    a %= mod;
    while(b) {
        if(b & 1LL) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

vector<ll> FactorialInv, Factorial;

void init_combination(ll MAX){
    Factorial.resize(MAX + 1);
    FactorialInv.resize(MAX + 1);
    Factorial[0] = 1;
    for(int i = 1; i <= MAX; i++){
        Factorial[i] = Factorial[i - 1] * i % mod;
    }
    FactorialInv[MAX] = power(Factorial[MAX], mod - 2);
    for(ll i = MAX - 1; i >= 0; i--) {
        FactorialInv[i] = FactorialInv[i+1] * (i+1) % mod;
    }
}

ll combination(ll a, ll b){
    if((a == b) || (b == 0)){
        return 1;
    }
    if(a < b) return 0;
    if(b < 0) return 0;
    ll ans = Factorial[a] * FactorialInv[b] % mod;
    ans = ans * FactorialInv[a - b] % mod;
    return ans;
}

//modの値の確認をすること
int main(){
    ll h,w;
    cin >> h >> w;
    vector<vector<ll> > dp(h,vector<ll>(w,INF)),g(h + 1,vector<ll>(w + 1,0ll));
    dp[0][0] = 0;
    rep0(i, h)rep0(j, w){
        char c;
        cin >> c;
        if(c == '#')g[i][j] = 1;
    }
    rep0(i, h)rep0(j, w){
        if(i != h-1) dp[i + 1][j] = min(dp[i + 1][j],dp[i][j] + (g[i][j] == g[i + 1][j]? 0:1));
        if(j != w-1) dp[i][j + 1] = min(dp[i][j + 1],dp[i][j] + (g[i][j] == g[i][j + 1]? 0:1));
    }
    ll ans = dp[h - 1][w - 1] + g[0][0] + g[h - 1][w - 1];
    pf(ans / 2);
    return 0;
}