Codeforces 426 B Sereja and Mirroring【暴力枚举+模拟】水题

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 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 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.

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

题目大意：

给你一个N*M大的矩阵，让你找到最小的矩阵b（x*M），使得其景象之后能够得到原矩阵。

比如样例1：

001

110

110

001

就可以通过第一个2*M的矩阵：

001

110

镜像翻转之后得到：

110

001

拼接之后得到：

001

110

110

001

所以答案是2、

思路：

那么从小到大暴力枚举这个x，使得对x*2^y==n然后模拟镜像翻转过程即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[150][150];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        if(n%2==1)
        {
            printf("%d\n",n);
            continue;
        }
        else
        {
            for(int i=1;i<=n;i++)
            {
                if(n%i==0)
                {
                    int ok=0;
                    int tmp=i;
                    for(int j=0;;j++)
                    {
                        if(tmp==n)
                        {
                            ok=1;break;
                        }
                        if(tmp>n)
                        {
                            ok=0;
                            break;
                        }
                        tmp*=2;
                    }
                    if(ok==0)continue;
                    ok=1;
                    int x=0,y=0,flag=1;
                    for(int j=0;j<n;j++)
                    {
                        for(int k=0;k<m;k++)
                        {
                            if(a[j][k]==a[x][y])
                            {
                                if(flag==1)
                                {
                                    y++;
                                    if(y>=m)y=0,x++;
                                    if(x==i)
                                    {
                                        flag=0;
                                        x=i-1;y=0;
                                    }
                                }
                                else
                                {
                                    y++;
                                    if(y>=m)y=0,x--;
                                    if(x==-1)
                                    {
                                        flag=1;
                                        x=0;y=0;
                                    }
                                }
                            }
                            else ok=0;
                        }
                    }
                    if(ok==1)
                    {
                        printf("%d\n",i);break;
                    }
                }
            }
        }
    }
}