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>
#define ll long long
#define uint unsigned
#define ull unsigned ll
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a))
#define CPY(a,b) memcpy(a,b,sizeof(a))
using namespace std;
const int N=200;
int n,m,fl[N];
char s[N][N];
int f[2][N][N][N];
int g[2][N][N][N];
int main(){
	scanf("%d%d",&n,&m);
	For(i,1,n){
		scanf("%s",s[i]+1);
		For(j,1,n) s[i][j]=(s[i][j]=='.'?0:1); 
	}
	For(i,1,n){
		For(j,1,m) fl[j]=s[i][j];
		For(j,i,n){
			For(k,1,m) fl[k]=(fl[k]!=s[j][k]?2:fl[k]);
			int las=fl[m],ed=m;
			f[0][i][j][m+1]=m+1;
			Rep(k,m,1){
				if (las!=fl[k]) las=fl[k],ed=k;
				f[0][i][j][k]=(las==2?k:ed+1);
			}
		}
	}
	For(i,1,m){
		For(j,1,n) fl[j]=s[j][i];
		For(j,i,m){
			For(k,1,n) fl[k]=(fl[k]!=s[k][j]?2:fl[k]);
			int las=fl[n],ed=n;
			g[0][i][j][n+1]=n+1;
			Rep(k,n,1){
				if (las!=fl[k]) las=fl[k],ed=k;
				g[0][i][j][k]=(las==2?k:ed+1);
			}
		}
	}
	int pre=0,suf=1;
	for (int ans=0;ans<=100;){
		//printf("%d %d\n",f[pre][1][n][1],f[pre][1][n][2]);
		if (f[pre][1][n][1]==m+1){
			printf("%d\n",ans);
			return 0;
		}
		For(i,1,n) For(j,i,n) For(k,1,m+1)
			f[suf][i][j][k]=f[pre][i][j][f[pre][i][j][k]];
		For(i,1,m) For(j,i,m) For(k,1,n+1)
			g[suf][i][j][k]=g[pre][i][j][g[pre][i][j][k]];
		For(i,1,n) For(j,i,n){
			int ed=1;
			For(k,1,m){
				ed=max(ed,k);
				for (;ed!=m+1&&g[pre][k][ed][g[pre][k][ed][i]]>j;++ed);
				f[suf][i][j][k]=max(f[suf][i][j][k],ed);
			}
		}
		For(i,1,m) For(j,i,m){
			int ed=1;
			For(k,1,n){
				ed=max(ed,k);
				for (;ed!=n+1&&f[pre][k][ed][f[pre][k][ed][i]]>j;++ed);
				g[suf][i][j][k]=max(g[suf][i][j][k],ed);
			}
		}
		++ans;
		swap(pre,suf);
	}
}
/*
f[i][j][k][l]:上界为i，下界为j，左界为k，答案为l的最右右界
g[i][j][k][l]:左界为i，右界为j，上界为k，答案为l的最下下界 
竖着划分可以xjb转移 
横着划分二分右边界检验 
*/