#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<ll>;
using vvi = vector<vi>;
const ll mod = 1000000007;
int n;
vvi g;

int go(int v) {
	vi d;
	d.assign(n, -1);
	int ans = 1;
	d[v] = 1;
	queue<int> q;
	q.push(v);
	int md;
	while (q.size()) {
		int V = q.front();
		q.pop();
		md = d[V];
		for (auto i : g[V]) {
			if(d[i] != -1) continue;
			d[i] = d[V]+1;
			q.push(i);
		}
//		cout << V
	}
	for(int i = 0; i < n; i++)
		for(auto j : g[i])
			if(abs(d[i]-d[j])!=1)
				return -1;
//	cout << v << " "
	return md;
}

int main() {
	ios::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);
	cin >> n;
	g.resize(n);
	char c;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cin >> c;
			if (c == '1')
				g[i].pb(j);
		}
	}
	int ans = -1;
	for (int i = 0; i < n; i++)
		ans = max(ans, go(i));
	cout << ans;
}

### 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 all(x) x.begin(), x.end()
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<ll>;
using vvi = vector<vi>;
const ll mod = 1000000007;
int n;
vvi g;

int go(int v) {
	vi d;
	d.assign(n, -1);
	int ans = 1;
	d[v] = 1;
	queue<int> q;
	q.push(v);
	int md;
	while (q.size()) {
		int V = q.front();
		q.pop();
		md = d[V];
		for (auto i : g[V]) {
			if(d[i] != -1) continue;
			d[i] = d[V]+1;
			q.push(i);
		}
//		cout << V
	}
	for(int i = 0; i < n; i++)
		for(auto j : g[i])
			if(abs(d[i]-d[j])!=1)
				return -1;
//	cout << v << " "
	return md;
}

int main() {
	ios::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);
	cin >> n;
	g.resize(n);
	char c;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cin >> c;
			if (c == '1')
				g[i].pb(j);
		}
	}
	int ans = -1;
	for (int i = 0; i < n; i++)
		ans = max(ans, go(i));
	cout << ans;
}

```