Problem:
Score : 500 points

Problem Statement

Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S.
If S_{i,j} is 1, there is an edge connecting Vertex i and j; otherwise, there is no such edge.

Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.

Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.

Constraints

2 \leq N \leq 200

S_{i,j} is 0 or 1.

S_{i,i} is 0.

S_{i,j}=S_{j,i}

The given graph is connected.

N is an integer.

Input

Input is given from Standard Input in the following format:

N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}

Output

If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1.
Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.

Sample Input 1

2
01
10

Sample Output 1

2

We can put Vertex 1 in V_1 and Vertex 2 in V_2.

Sample Input 2

3
011
101
110

Sample Output 2

-1

Sample Input 3

6
010110
101001
010100
101000
100000
010000

Sample Output 3

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define DEBUG if(0)
#define lli long long int
#define ldouble long double
using namespace std;

const int maxN = 301;
const lli inf = 1e18;
char s[maxN]; int n; lli k;

lli solve()
{
  lli ans = 0;
  for (int i = 0; i < n - 1; i ++)
    if (s[i] == s[i + 1])
      ans ++, i ++;
  return ans;
}

int main()
{
  scanf("%s %lld", s, &k);
  n = strlen(s); int originalSize = n;

  lli f[4]; f[0] = 0; f[1] = solve();
  for (int j = 1; j < 3; j ++)
  {
    for (int i = 0; i < originalSize; i ++)
      s[i + originalSize*j] = s[i];
    n += originalSize;
    f[j + 1] = solve();
  }

  if (k <= 3) printf("%lld\n", f[k]);
  else
  {
    lli ans = f[1];
    ans += ((k - 1) >> 1LL) * (f[3] - f[1]);
    if ((k - 1) & 1LL) ans += f[2] - f[1];
    printf("%lld\n", ans);
  }

  return 0;
}