#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
bool g[202][202];
char s[202][202];

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, ans = -1;
	cin >> N;
	rep(i,N)rep(j,N){
		cin >> s[i][j];
		if(s[i][j]=='1')g[i][j]=1;
	}
	
	rep(i,N){
		int cnt = 0;
		vector<bool> v(N);
		vector<vector<int>> V(N);
		v[i] = 1;
		V[0].push_back(i);
		bool ok = 1;
		while(ok){
			ok = 0;
			for(auto x:V[cnt]){
				rep(a,N){
					if(v[a]==0&&s[x][a]=='1'){
						V[cnt+1].push_back(a);
						v[a]=1;
						ok = 1;
					}
				}
			}
			cnt++;
		}
		ok = 1;
		for(int j=0;j<cnt;j++){
			for(int a=0;a<V[j].size();a++){
				for(int b=0;b<V[j].size();b++){
					int x = V[j][a];
					int y = V[j][b];
					if(g[x][y]){
						ok = 0;
						break;
					}
				}
				if(!ok)break;
			}
			if(!ok)break;
		}
		if(ok)ans = max(ans,cnt);
	}
	cout << ans << endl;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
bool g[202][202];
char s[202][202];

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, ans = -1;
	cin >> N;
	rep(i,N)rep(j,N){
		cin >> s[i][j];
		if(s[i][j]=='1')g[i][j]=1;
	}
	
	rep(i,N){
		int cnt = 0;
		vector<bool> v(N);
		vector<vector<int>> V(N);
		v[i] = 1;
		V[0].push_back(i);
		bool ok = 1;
		while(ok){
			ok = 0;
			for(auto x:V[cnt]){
				rep(a,N){
					if(v[a]==0&&s[x][a]=='1'){
						V[cnt+1].push_back(a);
						v[a]=1;
						ok = 1;
					}
				}
			}
			cnt++;
		}
		ok = 1;
		for(int j=0;j<cnt;j++){
			for(int a=0;a<V[j].size();a++){
				for(int b=0;b<V[j].size();b++){
					int x = V[j][a];
					int y = V[j][b];
					if(g[x][y]){
						ok = 0;
						break;
					}
				}
				if(!ok)break;
			}
			if(!ok)break;
		}
		if(ok)ans = max(ans,cnt);
	}
	cout << ans << endl;
}
```