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;
typedef long long ll;
const int N=190,p=1000000007,inv3=(p+1)/3;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int qpow(int x,int k){
	int t=1;
	for (;k;k>>=1){
		if (k&1) t=(ll)t*x%p;
		x=(ll)x*x%p;
	}
	return t;
}
int n,m,f[2][N][N][N],g[2][N][N][N];
char a[N][N];
bool check(int ans,int i,int j,int l,int r){
	int t=max(f[ans&1][i][l][r]+1,i);
	if (t>j || f[ans&1][t][l][r]>=j) return true;
	t=max(g[ans&1][l][i][j]+1,l);
	if (t>r || g[ans&1][t][i][j]>=r) return true;
	return false;
}
int main(){
	n=read();m=read();
	for (int i=1;i<=n;i++) scanf(" %s",a[i]+1);
	for (int i=n;i;i--)
	for (int l=1;l<=m;l++)
	for (int r=l;(r<=m)&&(a[i][l]==a[i][r]);r++)
		f[0][i][l][r]=(a[i][l]==a[i+1][l]) ? max(i,f[0][i+1][l][r]) : i;
	for (int i=m;i;i--)
	for (int l=1;l<=n;l++)
	for (int r=l;(r<=n)&&(a[l][i]==a[r][i]);r++)
		g[0][i][l][r]=(a[l][i]==a[l][i+1]) ? max(i,g[0][i+1][l][r]) : i;
	int ans;
	for (ans=0;f[ans&1][1][1][m]<n;ans++){
		for (int i=1;i<=n;i++)
		for (int l=1;l<=m;l++)
		for (int r=l;r<=m;r++){
			int cur=max(f[ans&1][i][l][r],i-1),k;
			for (k=0;(cur+(1<<k)<=n)&&check(ans,i,cur+(1<<k),l,r);k++);
			for (k--;k>=0;k--) if ((cur+(1<<k)<=n)&&check(ans,i,cur+(1<<k),l,r)) cur+=1<<k;
			f[(ans+1)&1][i][l][r]=cur;
		}
		for (int i=1;i<=m;i++)
		for (int l=1;l<=n;l++)
		for (int r=l;r<=n;r++){
			int cur=max(g[ans&1][i][l][r],i-1),k;
			for (k=0;(cur+(1<<k)<=m)&&check(ans,l,r,i,cur+(1<<k));k++);
			for (k--;k>=0;k--) if ((cur+(1<<k)<=m)&&check(ans,l,r,i,cur+(1<<k))) cur+=1<<k;
			g[(ans+1)&1][i][l][r]=cur;
		}
	}
	printf("%d\n",ans);
	return 0;
}