#include <bits/stdc++.h>
using namespace std;
const int INF = 1001001001;

int main(){
  int n;
  cin >> n;
  vector<string> s(n);
  for(int i=0; i<n; ++i) cin >> s[i];
  vector<int> oz(n, -1);
  oz[0] = 0;
  queue<int> que;
  que.push(0);
  while(!que.empty()){
    int x = que.front();
    que.pop();
    for(int i=0; i<n; ++i){
      if(s[x][i] == '0') continue;
      if(oz[i] == -1){
        oz[i] = (oz[x] ? 0:1);
        que.push(i);
      }
      else if(oz[i] == oz[x]){
        cout << -1 << endl;
        return 0;
      }
    }
  }
  vector<vector<int>> wf(n,vector<int>(n,INF));
  for(int i=0; i<n; ++i) for(int j=0; j<n; ++j){
    if(i == j) wf[i][j] = 0;
    else if(s[i][j] == '1') wf[i][j] = 1;
  }
  for(int k=0; k<n; ++k) for(int i=0; i<n; ++i) for(int j=0; j<n; ++j){
    wf[i][j] = min(wf[i][j], wf[i][k] + wf[k][j]);
  }
  int ans = 0;
  for(int i=0; i<n; ++i) for(int j=0; j<n; ++j){
    ans = max(ans,wf[i][j]);
  }
  cout << ans+1 << endl;
}
### Prompt

Your challenge is to write a cpp solution to 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 <bits/stdc++.h>
using namespace std;
const int INF = 1001001001;

int main(){
  int n;
  cin >> n;
  vector<string> s(n);
  for(int i=0; i<n; ++i) cin >> s[i];
  vector<int> oz(n, -1);
  oz[0] = 0;
  queue<int> que;
  que.push(0);
  while(!que.empty()){
    int x = que.front();
    que.pop();
    for(int i=0; i<n; ++i){
      if(s[x][i] == '0') continue;
      if(oz[i] == -1){
        oz[i] = (oz[x] ? 0:1);
        que.push(i);
      }
      else if(oz[i] == oz[x]){
        cout << -1 << endl;
        return 0;
      }
    }
  }
  vector<vector<int>> wf(n,vector<int>(n,INF));
  for(int i=0; i<n; ++i) for(int j=0; j<n; ++j){
    if(i == j) wf[i][j] = 0;
    else if(s[i][j] == '1') wf[i][j] = 1;
  }
  for(int k=0; k<n; ++k) for(int i=0; i<n; ++i) for(int j=0; j<n; ++j){
    wf[i][j] = min(wf[i][j], wf[i][k] + wf[k][j]);
  }
  int ans = 0;
  for(int i=0; i<n; ++i) for(int j=0; j<n; ++j){
    ans = max(ans,wf[i][j]);
  }
  cout << ans+1 << endl;
}
```