#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define MAX 1<<29
typedef long long ll;


vector<vector<ll>> g;
vector<ll> depth;
bool flag = false;

void dfs(int pos, int dep){
  if(depth[pos] >= 0) {
      if(depth[pos] != dep) flag = true;
      return;
  }
  depth[pos] = dep;
  for(auto x:g[pos]) dfs(x,dep^1);
}

int main(){
  ll n;cin>>n;
  vector<string> s(n);
  rep(i,n) cin>>s[i];
  vector<vector<ll>> dist(n, vector<ll>(n, MAX));
  g.resize(n);
  rep(i,n){
    rep(j,n) if(s[i][j] == '1') {
        g[i].push_back(j);
        dist[i][j] = 1;
    }
  }
  //二部グラフ判定
  depth.resize(n, 0);
  rep(i,n) depth[i] = -1;
  dfs(0,0);
  if(flag){
      puts("-1");
      return 0;
  }
  //warshall_floyd法
  rep(i,n) dist[i][i]=0;
  rep(k,n)rep(i,n)rep(j,n) dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]);
  ll ans=0;
  rep(i,n) ans = max(ans,*max_element(dist[i].begin(),dist[i].end()));
  cout << ans + 1 << endl;
  return 0;
}
### Prompt

Generate 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;
#define rep(i,n) for(int i=0;i<n;i++)
#define MAX 1<<29
typedef long long ll;


vector<vector<ll>> g;
vector<ll> depth;
bool flag = false;

void dfs(int pos, int dep){
  if(depth[pos] >= 0) {
      if(depth[pos] != dep) flag = true;
      return;
  }
  depth[pos] = dep;
  for(auto x:g[pos]) dfs(x,dep^1);
}

int main(){
  ll n;cin>>n;
  vector<string> s(n);
  rep(i,n) cin>>s[i];
  vector<vector<ll>> dist(n, vector<ll>(n, MAX));
  g.resize(n);
  rep(i,n){
    rep(j,n) if(s[i][j] == '1') {
        g[i].push_back(j);
        dist[i][j] = 1;
    }
  }
  //二部グラフ判定
  depth.resize(n, 0);
  rep(i,n) depth[i] = -1;
  dfs(0,0);
  if(flag){
      puts("-1");
      return 0;
  }
  //warshall_floyd法
  rep(i,n) dist[i][i]=0;
  rep(k,n)rep(i,n)rep(j,n) dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]);
  ll ans=0;
  rep(i,n) ans = max(ans,*max_element(dist[i].begin(),dist[i].end()));
  cout << ans + 1 << endl;
  return 0;
}
```