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 a2x × 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 30 0 11 1 01 1 00 0 1

output

2

input

3 30 0 00 0 00 0 0

output

3

input

8 101100110

output

2

Note

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

001110

If we perform a mirroring operation with this matrix, we get the matrix a that is given in the input:

001110110001

解题说明：此题可以等价看成一道图形题，按照对称型的关系来加以判断。为了找到通过不断对称构成整个图形的最小单位，首先应该判断全部图形是否构成对称，如果对称则判断其一半图形是否对称，直到遇到不构成对称的情况，这时就找到了最小单元。

#include<iostream>#include<cstdio>#include<cmath>#include<cstdlib>#include <algorithm>#include<cstring>#include<string>using namespace std;int a[100][100], b[100][100], c[100][100];int main(){int n, m, i, j, line, temp = 0;scanf("%d%d", &n, &m);for (i = 0; i < n; i++){for (j = 0; j < m; j++){scanf("%d", &a[i][j]);}}line = n;while (!(line % 2)){for (i = 0; i < line / 2; i++){for (j = 0; j < m; j++){if (a[i][j] != a[line - 1 - i][j]){temp++;}}}if (temp){break;}else{line /= 2;}}printf("%d\n", line);return 0;}