A brick is defined as a rectangle with integer side lengths with either width 1 or height 1 (or both).

There is an n× m grid, and each cell is colored either black or white. A tiling is a way to place bricks onto the grid such that each black cell is covered by exactly one brick, and each white cell is not covered by any brick. In other words, bricks are placed on black cells only, cover all black cells, and no two bricks overlap.

<image> An example tiling of the first test case using 5 bricks. It is possible to do better, using only 4 bricks. 

What is the minimum number of bricks required to make a valid tiling?

Input

The first line contains two integers n, m (1≤ n, m≤ 200) — the number of rows and columns, respectively.

The next n lines describe the grid. The i-th of these lines contains a string of length m, where the j-th character denotes the color of the cell in row i, column j. A black cell is given by "#", and a white cell is given by ".".

It is guaranteed that there is at least one black cell.

Output

Output a single integer, the minimum number of bricks required.

Examples

Input


3 4
#.##
####
##..


Output


4


Input


6 6
######
##....
######
##...#
##...#
######


Output


6


Input


10 8
####..##
#..#.##.
#..#.###
####.#.#
....####
.###.###
###.#..#
########
###..###
.##.###.


Output


18

Note

The first test case can be tiled with 4 bricks placed vertically.

The third test case can be tiled with 18 bricks like this:

<image>
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
void ckmn(T &a, T b) {
  a = min(a, b);
}
template <typename T>
void ckmx(T &a, T b) {
  a = max(a, b);
}
void rd(int &x) { scanf("%i", &x); }
void rd(long long &x) { scanf("%lld", &x); }
void rd(char *x) { scanf("%s", x); }
void rd(double &x) { scanf("%lf", &x); }
void rd(string &x) { scanf("%s", &x); }
template <typename T1, typename T2>
void rd(pair<T1, T2> &x) {
  rd(x.first);
  rd(x.second);
}
template <typename T>
void rd(vector<T> &x) {
  for (T &i : x) rd(i);
}
template <typename T, typename... A>
void rd(T &x, A &...args) {
  rd(x);
  rd(args...);
}
template <typename T>
void rd() {
  T x;
  rd(x);
  return x;
}
int ri() {
  int x;
  rd(x);
  return x;
}
template <typename T>
vector<T> rv(int n) {
  vector<T> x(n);
  rd(x);
  return x;
}
template <typename T>
void ra(T a[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]);
}
template <typename T1, typename T2>
void ra(T1 a[], T2 b[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]);
}
template <typename T1, typename T2, typename T3>
void ra(T1 a[], T2 b[], T3 c[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]), rd(c[st + i]);
}
void re(vector<int> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    rd(u, v);
    E[u].push_back(v);
    if (!dir) E[v].push_back(u);
  }
}
template <typename T>
void re(vector<pair<int, T>> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    T w;
    rd(u, v, w);
    E[u].push_back({v, w});
    if (!dir) E[v].push_back({u, w});
  }
}
template <int NSZ, int ESZ, typename C, C inf>
struct Dinic {
  struct Edge {
    int v, go;
    C c;
    Edge(int _v, C _c, int _go) : v(_v), c(_c), go(_go) {}
    Edge() {}
  };
  Edge edges[ESZ * 2];
  int G[NSZ], ptr[NSZ];
  int dist[NSZ], q[NSZ], ql, qr;
  int s, t, n, xb;
  void init(int _n) {
    n = _n;
    xb = ql = qr = 0;
    for (int i = 0; i <= n; i++) G[i] = -1;
  }
  Dinic(int _n) { init(_n); }
  Dinic() {}
  void AddEdge(int u, int v, C c) {
    edges[xb] = Edge(v, c, G[u]);
    G[u] = xb++;
    edges[xb] = Edge(u, 0, G[v]);
    G[v] = xb++;
  }
  C DFS(int u, C flow) {
    if (u == t) return flow;
    if (dist[u] >= dist[t]) return 0;
    C ans = 0;
    for (int &e = ptr[u]; e != -1; e = edges[e].go) {
      if (edges[e].c < 0 || dist[edges[e].v] != dist[u] + 1) continue;
      C cut = DFS(edges[e].v, min(flow - ans, edges[e].c));
      ans += cut;
      edges[e].c -= cut;
      edges[e ^ 1].c += cut;
      if (flow == ans) break;
    }
    return ans;
  }
  bool BFS() {
    for (int i = 0; i <= n; ++i) dist[i] = -1, ptr[i] = G[i];
    dist[s] = 0;
    ql = qr = 0;
    q[qr++] = s;
    while (ql != qr) {
      int u = q[ql++];
      for (int e = G[u]; e != -1; e = edges[e].go) {
        if (edges[e].c > 0 && dist[edges[e].v] == -1) {
          dist[edges[e].v] = dist[u] + 1;
          q[qr++] = edges[e].v;
        }
      }
    }
    return dist[t] != -1;
  }
  C MaxFlow(int _s, int _t) {
    s = _s;
    t = _t;
    C Flow = 0;
    while (BFS()) Flow += DFS(s, inf);
    return Flow;
  }
  vector<bool> vis;
  void REC(int u) {
    vis[u] = 1;
    for (int e = G[u]; e != -1; e = edges[e].go)
      if (edges[e].c > 0 && !vis[edges[e].v]) REC(edges[e].v);
  }
  vector<bool> MinCut() {
    vis.assign(n + 1, false);
    REC(s);
    return vis;
  }
};
const int inf = 1e9 + 7;
const int N = 205;
char base[N][N];
int main() {
  int n, m;
  rd(n, m);
  for (int i = 1; i <= n; i++) rd(base[i] + 1);
  int s = n * m * 2 + 1, t = n * m * 2 + 2;
  auto idx1 = [&](int i, int j) { return (i - 1) * m + j; };
  auto idx2 = [&](int i, int j) { return (i - 1) * m + j + n * m; };
  Dinic<N * N * 2, N * N * 6, int, inf> Flow(t);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++)
      if (base[i][j] == '#') {
        cnt++;
        if (base[i + 1][j] == '#' && base[i][j + 1] == '#')
          Flow.AddEdge(idx1(i, j), idx2(i, j), 1);
        if (base[i - 1][j] == '#' && base[i][j + 1] == '#')
          Flow.AddEdge(idx1(i - 1, j), idx2(i, j), 1);
        if (base[i + 1][j] == '#' && base[i][j - 1] == '#')
          Flow.AddEdge(idx1(i, j), idx2(i, j - 1), 1);
        if (base[i - 1][j] == '#' && base[i][j - 1] == '#')
          Flow.AddEdge(idx1(i - 1, j), idx2(i, j - 1), 1);
        if (base[i + 1][j] == '#') Flow.AddEdge(s, idx1(i, j), 1), cnt--;
        if (base[i][j + 1] == '#') Flow.AddEdge(idx2(i, j), t, 1), cnt--;
      }
  }
  printf("%i\n", cnt + Flow.MaxFlow(s, t));
  return 0;
}
