#include <iostream>
#include <stack>

using namespace std;

#define rep(i,n) for(int i=0;i<(n);++i)
#define out(v) cout<<v<<"\n"
const int INF = 1e6;

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int N; cin >> N;

  bool edge[N][N];
  rep(i, N) rep(j, N) {
    char c; cin >> c;
    edge[i][j] = (c == '1');
  }

  int d[N][N];
  rep(i, N) rep(j, N) d[i][j] = edge[i][j] ? 1 : INF;
  rep(k, N) rep(i, N) rep(j, N) {
    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
  }

  int color[N];
  rep(i, N) color[i] = INF;
  color[0] = 0;

  stack<int> st;
  st.push(0);
  while (!st.empty()) {
    int p = st.top(); st.pop();
    rep(i, N) {
      if (!edge[i][p])
        continue;
      if (color[i] == color[p]) {
        out(-1);
        return 0;
      }
      if (color[i] == INF) {
        color[i] = 1 - color[p];
        st.push(i);
      }
    }
  }

  int r = 0;
  rep(i, N) rep(j, N) {
    if (i == j)
      continue;
    r = max(r, d[i][j]);
  }
  out(r+1);
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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.

Examples

Input

2
01
10


Output

2


Input

3
011
101
110


Output

-1


Input

6
010110
101001
010100
101000
100000
010000


Output

4

### Response

```cpp
#include <iostream>
#include <stack>

using namespace std;

#define rep(i,n) for(int i=0;i<(n);++i)
#define out(v) cout<<v<<"\n"
const int INF = 1e6;

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int N; cin >> N;

  bool edge[N][N];
  rep(i, N) rep(j, N) {
    char c; cin >> c;
    edge[i][j] = (c == '1');
  }

  int d[N][N];
  rep(i, N) rep(j, N) d[i][j] = edge[i][j] ? 1 : INF;
  rep(k, N) rep(i, N) rep(j, N) {
    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
  }

  int color[N];
  rep(i, N) color[i] = INF;
  color[0] = 0;

  stack<int> st;
  st.push(0);
  while (!st.empty()) {
    int p = st.top(); st.pop();
    rep(i, N) {
      if (!edge[i][p])
        continue;
      if (color[i] == color[p]) {
        out(-1);
        return 0;
      }
      if (color[i] == INF) {
        color[i] = 1 - color[p];
        st.push(i);
      }
    }
  }

  int r = 0;
  rep(i, N) rep(j, N) {
    if (i == j)
      continue;
    r = max(r, d[i][j]);
  }
  out(r+1);
  return 0;
}

```