Problem:
Score : 1000 points

Problem Statement

Note the unusual memory limit.

For a rectangular grid where each square is painted white or black, we define its complexity as follows:

If all the squares are black or all the squares are white, the complexity is 0.

Otherwise, divide the grid into two subgrids by a line parallel to one of the sides of the grid, and let c_1 and c_2 be the complexities of the subgrids. There can be multiple ways to perform the division, and let m be the minimum value of \max(c_1, c_2) in those divisions. The complexity of the grid is m+1.

You are given a grid with H horizontal rows and W vertical columns where each square is painted white or black.
HW characters from A_{11} to A_{HW} represent the colors of the squares.
A_{ij} is # if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is . if that square is white.

Find the complexity of the given grid.

Constraints

1 \leq H,W \leq 185

A_{ij} is # or ..

Input

Input is given from Standard Input in the following format:

H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}

Output

Print the complexity of the given grid.

Sample Input 1

3 3
...
.##
.##

Sample Output 1

2

Let us divide the grid by the boundary line between the first and second columns.
The subgrid consisting of the first column has the complexity of 0, and the subgrid consisting of the second and third columns has the complexity of 1, so the whole grid has the complexity of at most 2.

Sample Input 2

6 7
.####.#
#....#.
#....#.
#....#.
.####.#
#....##

Sample Output 2

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define trep(i,s,t) for (int i = (t)-1; i >= s; --i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("YES");}else{puts("NO");}
const int MX = 186;
const int M = 9;

char s[MX][MX];
char d[2][MX][MX][MX];
char e[2][MX][MX][MX];

int main() {
  int h,w;
  scanf("%d%d",&h,&w);
  rep(i,h) scanf("%s",s[i]);
  int o = 0, p = 1;
  rep(si,h)rep(sj,w) {
    char c = s[si][sj];
    int tj = sj;
    trep(ti,si,h) {
      while (tj < w) {
        bool ok = true;
        srep(i,si,ti+1) if (s[i][tj] != c) { ok = false; break;}
        if (!ok) break;
        ++tj;
      }
      d[o][si][sj][ti] = tj;
    }
  }
  rep(si,h)rep(sj,w) {
    char c = s[si][sj];
    int ti = si;
    trep(tj,sj,w) {
      while (ti < h) {
        bool ok = true;
        srep(j,sj,tj+1) if (s[ti][j] != c) { ok = false; break;}
        if (!ok) break;
        ++ti;
      }
      e[o][si][sj][tj] = ti;
    }
  }
  rep(ans,15) {
    // rep(si,h)rep(sj,w)srep(ti,si,h) printf("%d %d %d %d\n",si,sj,ti,d[o][si][sj][ti]);
    // puts("---");
    // rep(si,h)rep(sj,w)srep(tj,sj,w) printf("%d %d %d %d\n",si,sj,tj,e[o][si][sj][tj]);
    // puts("===");
    if (d[o][0][0][h-1] >= w) {
      cout<<ans<<endl;
      return 0;
    }
    swap(o,p);
    rep(si,h)rep(sj,w)rep(ti,h) d[o][si][sj][ti] = d[p][si][sj][ti];
    rep(si,h)rep(sj,w)rep(tj,w) e[o][si][sj][tj] = e[p][si][sj][tj];
    rep(si,h)rep(sj,w) {
      srep(ti,si,h) {
        int tj = d[p][si][sj][ti]-1;
        while (tj >= sj) {
          if (tj+1 < w) {
            char nj = d[p][si][tj+1][ti];
            maxs(d[o][si][sj][ti],nj);
            maxs(e[o][si][sj][nj-1],char(ti+1));
          }
          if (ti+1 < h) {
            char ni = e[p][ti+1][sj][tj];
            maxs(d[o][si][sj][ni-1],char(tj+1));
            maxs(e[o][si][sj][tj],ni);
          }
          --tj;
        }
      }
      trep(ti,si,h-1) maxs(d[o][si][sj][ti], d[o][si][sj][ti+1]);
      trep(tj,sj,w-1) maxs(e[o][si][sj][tj], e[o][si][sj][tj+1]);
    }
  }
  return 0;
}