Let's assume that we are given a matrix b of size x × y, let's determine the operation of mirroring matrix b. The mirroring of matrix b is a 2x × y matrix c which has the following properties:

  * the upper half of matrix c (rows with numbers from 1 to x) exactly matches b; 
  * the lower half of matrix c (rows with numbers from x + 1 to 2x) is symmetric to the upper one; the symmetry line is the line that separates two halves (the line that goes in the middle, between rows x and x + 1). 



Sereja has an n × m matrix a. He wants to find such matrix b, that it can be transformed into matrix a, if we'll perform on it several (possibly zero) mirrorings. What minimum number of rows can such matrix contain?

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 100). Each of the next n lines contains m integers — the elements of matrix a. The i-th line contains integers ai1, ai2, ..., aim (0 ≤ aij ≤ 1) — the i-th row of the matrix a.

Output

In the single line, print the answer to the problem — the minimum number of rows of matrix b.

Examples

Input

4 3
0 0 1
1 1 0
1 1 0
0 0 1


Output

2


Input

3 3
0 0 0
0 0 0
0 0 0


Output

3


Input

8 1
0
1
1
0
0
1
1
0


Output

2

Note

In the first test sample the answer is a 2 × 3 matrix b:
    
    
      
    001  
    110  
    

If we perform a mirroring operation with this matrix, we get the matrix a that is given in the input:
    
    
      
    001  
    110  
    110  
    001  
    
/*
 * Remember a 7.0 student can know more than a 10.0 student.
 * Grades don't determine intelligence, they test obedience. 
 * I Never Give Up.
 */
import java.util.*;
import java.io.*;

import sun.java2d.pipe.OutlineTextRenderer;
//import javafx.util.Pair;
import static java.lang.System.out;
import static java.util.Arrays.*;
import static java.lang.Math.*;
public class ContestMain {
	private static Reader in=new Reader();
	private static StringBuilder ans=new StringBuilder();
	private static long MOD=1000000007;//10^9+7 
	private static final int N=10000000;
//	private static final double PI=3.1415;
//	private static ArrayList<Integer> v[]=new ArrayList[N];
//	private static int color[]=new int[N]; //For Graph Coloring
//	private static boolean mark[]=new boolean[N];
//	private static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
//	private static void dfs(int node){mark[node]=true;for(int x:v[node]){if(!mark[x]){dfs(x);}}}
	private static long powmod(long x,long n,long m){
		if(n==0)return 1;
		else if(n%2==0)return(powmod((x*x)%m,n/2,m));
		else return (x*(powmod((x*x)%m,(n-1)/2,m)))%m;
	}
	public static void solve(String ar[],int n){
		if(n%2!=0){
			out.println(n);
		}
		else{
			int i=0;
			while(n%2==0){
				for(i=0;i<n;i++){
					if(!ar[i].equals(ar[n-1-i]))break;
				}
				if(i<n)break;
				else n=n/2;
			}
			out.println(n);
		}
		
	}
	public static void main(String[] args) throws IOException{
		int n=in.nextInt();
		int m=in.nextInt();
		String ar[]=new String[n];
		for(int i=0;i<n;i++)
			ar[i]=in.nextLine();
		solve(ar,n);
	}
	static class Reader{
        BufferedReader br;
        StringTokenizer st;
 
        public Reader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}  