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<cstdio>
#include<algorithm>
using namespace std;
const int N=190;
int n,m,f[N][N][N],g[N][N][N],ff[N][N][N],gg[N][N][N];
char str[N][N];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)scanf("%s",str[i]+1);
	for(int i=n;i;--i)
		for(int j=1;j<=m;++j){
			f[i][j][j]=(str[i][j]==str[i+1][j]?f[i+1][j][j]:i+1);
			if(str[i][j]==str[i][j-1])
				for(int k=1;k<j;++k)
					f[i][k][j]=min(f[i][k][j-1],f[i][j][j]);
			else
				for(int k=1;k<j;++k)
					f[i][k][j]=i;
		}
	for(int i=m;i;--i)
		for(int j=1;j<=n;++j){
			g[i][j][j]=(str[j][i]==str[j][i+1]?g[i+1][j][j]:i+1);
			if(str[j][i]==str[j-1][i])
				for(int k=1;k<j;++k)
					g[i][k][j]=min(g[i][k][j-1],g[i][j][j]);
			else
				for(int k=1;k<j;++k)
					g[i][k][j]=i;
		}
	for(int ans=0;;++ans){
		if(f[1][1][m]==n+1)
			return printf("%d\n",ans),0;
		swap(f,ff);swap(g,gg);
		for(int i=1;i<=n;++i)
			for(int j=1;j<=m;++j)
				for(int k=j;k<=m;++k)
					if(ff[i][j][k]==n+1)
						f[i][j][k]=n+1;
					else{
						f[i][j][k]=ff[ff[i][j][k]][j][k];
						while(f[i][j][k]<=n&&gg[gg[j][i][f[i][j][k]]][i][f[i][j][k]]>k)++f[i][j][k];
					}
		for(int i=1;i<=m;++i)
			for(int j=1;j<=n;++j)
				for(int k=j;k<=n;++k)
					if(gg[i][j][k]==m+1)
						g[i][j][k]=m+1;
					else{
						g[i][j][k]=gg[gg[i][j][k]][j][k];
						while(g[i][j][k]<=m&&ff[ff[j][i][g[i][j][k]]][i][g[i][j][k]]>k)++g[i][j][k];
					}
	}
}