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  
    
#include <bits/stdc++.h>
using namespace std;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
void fast() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
}
void online_judge() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
const int flag_max = 0x3f3f3f3f;
const long long OO = 2e9 + 10;
const double EPS = (1e-8);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
const long long MOD = 1e14 + 7;
long long Power(long long n, long long deg) {
  if (!deg)
    return 1;
  else if (deg & 1)
    return (Power(n, deg - 1) * n);
  else {
    long long half = Power(n, deg / 2);
    return ((half * half));
  }
}
int n, m;
vector<string> row;
void mirrorings(int sz, vector<string>& now) {
  if (sz == n) return;
  for (int i = now.size() - 1; i >= 0; i--) now.push_back(now[i]);
  if (n % (int)now.size() == 0) mirrorings(2 * sz, now);
}
bool check(vector<string> carry) {
  for (int i = 0; i < row.size(); i++)
    if (row[i] != carry[i]) return 0;
  return 1;
}
int main() {
  fast();
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    string now = "";
    for (int j = 0; j < m; j++) {
      char X;
      cin >> X;
      now += X;
    }
    row.push_back(now);
  }
  int ans = n;
  vector<string> flag;
  for (int i = 0; i < n; i++) {
    flag.push_back(row[i]);
    vector<string> now = flag;
    if (n % (int)flag.size() == 0) {
      mirrorings(flag.size(), now);
      if (check(now)) ans = i + 1, i = n;
    }
  }
  cout << ans;
  return 0;
}
