Codeforces Sereja and Mirroring

Description

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 matrixa. The i-th line contains integers a_i1, a_i2, ..., a_im(0 ≤ a_ij ≤ 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 Input

Input

Output

Input

Output

Input

Output

Hint

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:

题目意思：给你一个矩阵a，如果是严格对称的，必须是偶对称，就消去下半部分，问最后剩多少部分.

利用字符串，方便比较。

#include <iostream>
#include <stdio.h>
#include <cstdlib>
using namespace std;
int main()
{int n,m,flag;string s[101];char ch;while (cin>>n>>m){flag=0;for (int i=0; i<n; i++){for (int j=0; j<m; j++){cin>>ch;if(j==0){s[i]=ch;}else{s[i]+=ch;}}}if(n%2){cout<<n<<endl;}else{while (!(n%2)){int i;for (i=0; i<n/2; i++){if(s[i]!=s[n-i-1]){break;}}if(i==n/2){n/=2;}else{cout<<n<<endl;flag=1;break;}}if(!flag){cout<<n<<endl;}}}return 0;
}

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