Problem: 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.

Examples

Input

3 3
...
.##
.##


Output

2


Input

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


Output

4

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int Mx(int a,int b){return a>b?a:b;}
int Mn(int a,int b){return a<b?a:b;}
const int N=190;
int n,m,sm[N][N],dp[2][N][N][N];
char ch[N][N];
bool chk(int x1,int y1,int x2,int y2)
{
  int s=sm[x2][y2]-sm[x2][y1-1]-sm[x1-1][y2]+sm[x1-1][y1-1];
  return (!s)||s==(x2-x1+1)*(y2-y1+1);
}
int main()
{
  scanf("%d%d",&n,&m);
  for(int i=1;i<=n;i++)
    scanf("%s",ch[i]+1);
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)sm[i][j]=sm[i][j-1]+(ch[i][j]=='.');
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)sm[i][j]+=sm[i-1][j];
  for(int i=1;i<=n;i++)for(int j1=1;j1<=m;j1++)for(int j2=j1;j2<=m;j2++)
  {
    int l=i,r=n,ans=i-1;
    while(l<=r)
    {
      int mid=l+r>>1;
      if(chk(i,j1,mid,j2))ans=mid,l=mid+1;else r=mid-1;
    }
    dp[0][i][j1][j2]=ans;
  }
  if(dp[0][1][1][m]==n){puts("0");return 0;}
  bool u=1,v=0;
  for(int t=1;t<=16;t++,swap(u,v))
  {
    for(int i=1;i<=n;i++)
      for(int j1=1;j1<=m;j1++)
        for(int j2=j1;j2<=m;j2++)
        {
          int d=dp[v][dp[v][i][j1][j2]+1][j1][j2];
          if(dp[v][i][j1][j2]==n)
          { dp[u][i][j1][j2]=n;continue;}
          int l=j1,r=j2-1;
          while(l<=r)
          {
            int mid=l+r>>1;
            int t1=dp[v][i][j1][mid],t2=dp[v][i][mid+1][j2];
            d=Mx(d,Mn(t1,t2));
            if(t1>=t2)l=mid+1; else r=mid-1;
          }
          dp[u][i][j1][j2]=d;
        }
    if(dp[u][1][1][m]==n){printf("%d\n",t);return 0;}
  }
  return 0;
}