#include <bits/stdc++.h>
using namespace std;
int main(){
	int n; cin >> n;
	vector<string> grid(n);
	for(int i = 0; i < n; i++) cin >> grid[i];
	int ans = -1;
	for(int i = 0; i < n; i++){
		vector<int> dist(n, -1);
		queue<int> q; q.push(i); dist[i] = 0;
		while(!q.empty()){
			int u = q.front(); q.pop();
			for(int j = 0; j < n; j++){
				if(grid[u][j] == '1' && dist[j] == -1){
					dist[j] = dist[u] + 1;
					q.push(j);
				}
			}
		}
		bool bad = false;
		for(int j = 0; j < n; j++){
			for(int k = 0; k < n; k++){
				if(grid[j][k] == '1' && abs(dist[j] - dist[k]) != 1){
					bad = true;	
				}	
			}
		}
		if(!bad){
			ans = max(ans, 1 + *max_element(dist.begin(), dist.end()));
		}
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Please formulate 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;
int main(){
	int n; cin >> n;
	vector<string> grid(n);
	for(int i = 0; i < n; i++) cin >> grid[i];
	int ans = -1;
	for(int i = 0; i < n; i++){
		vector<int> dist(n, -1);
		queue<int> q; q.push(i); dist[i] = 0;
		while(!q.empty()){
			int u = q.front(); q.pop();
			for(int j = 0; j < n; j++){
				if(grid[u][j] == '1' && dist[j] == -1){
					dist[j] = dist[u] + 1;
					q.push(j);
				}
			}
		}
		bool bad = false;
		for(int j = 0; j < n; j++){
			for(int k = 0; k < n; k++){
				if(grid[j][k] == '1' && abs(dist[j] - dist[k]) != 1){
					bad = true;	
				}	
			}
		}
		if(!bad){
			ans = max(ans, 1 + *max_element(dist.begin(), dist.end()));
		}
	}
	cout << ans << endl;
	return 0;
}

```