#include <bits/stdc++.h>

#define itn int
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())

using namespace std;

inline int nxt() {
	int x;
	scanf("%d", &x);
	return x;
}


int main() {
	int n = nxt();
	vector<vector<int>> a(n);
	for (int i = 0; i < n; ++i) {
		string s;
		cin >> s;
		for (int j = 0; j < n; ++j) {
			if (s[j] == '1') {
				a[i].push_back(j);
			}
		}
	}

	int ans = -1;
	for (int i = 0; i < n; ++i) {
		queue<int> q;
		q.push(i);
		vector<int> d(n, -1);
		d[i] = 0;
		while (!q.empty()) {
			int v = q.front();
			q.pop();
			for (int x : a[v]) {
				if (d[x] == -1) {
					d[x] = d[v] + 1;
					q.push(x);
				} else if (d[x] == d[v]) {
					cout << "-1\n";
					return 0;
				}
			}
		}
		ans = max(ans, *max_element(all(d)) + 1);
	}

	cout << ans << "\n";

	return 0;
}

### 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 itn int
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())

using namespace std;

inline int nxt() {
	int x;
	scanf("%d", &x);
	return x;
}


int main() {
	int n = nxt();
	vector<vector<int>> a(n);
	for (int i = 0; i < n; ++i) {
		string s;
		cin >> s;
		for (int j = 0; j < n; ++j) {
			if (s[j] == '1') {
				a[i].push_back(j);
			}
		}
	}

	int ans = -1;
	for (int i = 0; i < n; ++i) {
		queue<int> q;
		q.push(i);
		vector<int> d(n, -1);
		d[i] = 0;
		while (!q.empty()) {
			int v = q.front();
			q.pop();
			for (int x : a[v]) {
				if (d[x] == -1) {
					d[x] = d[v] + 1;
					q.push(x);
				} else if (d[x] == d[v]) {
					cout << "-1\n";
					return 0;
				}
			}
		}
		ans = max(ans, *max_element(all(d)) + 1);
	}

	cout << ans << "\n";

	return 0;
}

```