#include<bits/stdc++.h>
using namespace std;

const int N = 205;
typedef long long LL;

char ch;
int a[N][N], n, clr[N], ok, dis[N][N], ans;

void Dfs(int u, int col) {
	clr[u] = col;
	for(int i = 1; i <= n; i++)
		if(a[u][i] == 1) {
			if(clr[i]) {
				if(clr[i] == col) ok = 1;   
				continue;
			}
			Dfs(i, 3 - col);
		}
}

int main() {
	cin>>n;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) {
			cin>>ch;
			if(ch == '1') a[i][j] = 1;
			else {
				if(i == j) a[i][j] = 0;
				else a[i][j] = 1e9;
			}
		}
	Dfs(1, 1);
	if(ok) {
		cout<<-1<<endl;
		return 0;
	}
	for(int k = 1; k <= n; k++)
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				if(a[i][k] + a[k][j] < a[i][j]) 
					a[i][j] = a[i][k] + a[k][j];
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			ans = max(ans, a[i][j]);
	cout<<ans + 1<<endl;
	return 0;
}
### Prompt

Please create a solution in CPP 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 N = 205;
typedef long long LL;

char ch;
int a[N][N], n, clr[N], ok, dis[N][N], ans;

void Dfs(int u, int col) {
	clr[u] = col;
	for(int i = 1; i <= n; i++)
		if(a[u][i] == 1) {
			if(clr[i]) {
				if(clr[i] == col) ok = 1;   
				continue;
			}
			Dfs(i, 3 - col);
		}
}

int main() {
	cin>>n;
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++) {
			cin>>ch;
			if(ch == '1') a[i][j] = 1;
			else {
				if(i == j) a[i][j] = 0;
				else a[i][j] = 1e9;
			}
		}
	Dfs(1, 1);
	if(ok) {
		cout<<-1<<endl;
		return 0;
	}
	for(int k = 1; k <= n; k++)
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				if(a[i][k] + a[k][j] < a[i][j]) 
					a[i][j] = a[i][k] + a[k][j];
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			ans = max(ans, a[i][j]);
	cout<<ans + 1<<endl;
	return 0;
}
```