B. Sereja and Mirroring



time limit per test

1 second


memory limit per test

256 megabytes


input

standard input


output

standard output


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 a2 x × 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.



Sample test(s)




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







#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int num[111][111];

int main ()
{
    int n,m;
    scanf ("%d%d",&n,&m);

    int i,k;

    for (i = 0;i < n;i++)
        for (k = 0;k < m;k++)
            scanf ("%d",&num[i][k]);

    int ans = n;a

    if (n % 2)
        printf ("%d\n",n);
    else
    {
        int tn = n;
        while (1)
        {
            int tf = 1;
            for (i = 0;i < tn / 2;i++)
                for (k = 0;k < m;k++)
                    if (num[i][k] != num[tn - 1 - i][k])
                        tf = 0;
            if (tf)
            {
                if (tn % 2)
                    break;
                tn /= 2;
            }else
            {
                //tn *= 2;
                break;
            }
        }
        printf ("%d\n",tn);
    }

    return 0;
}