#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define all(x) (x).begin(), (x).end()

int q[200];
vector<vector<int>> ar;

int calc(int start) {
	int beg = 0;
	int end = 1;
	q[0] = start;
	int mark[200] = {};
	mark[start] = 1;
	while (beg < end) {
		int cur = q[beg++];
		for (int to: ar[cur]) {
			if (mark[to]) {
				if (abs(mark[to] - mark[cur]) != 1) {
					return -1;
				}
				continue;
			}
			q[end++] = to;
			mark[to] = mark[cur] + 1;
		}
	}
	return mark[q[end - 1]];
}

int main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	ar.resize(n);
	for (int i = 0; i < n; ++i) {
		string s;
		cin >> s;
		for (int j = i + 1; j < n; ++j) {
			if (s[j] == '1') {
				ar[i].push_back(j);
				ar[j].push_back(i);
			}
		}
	}
	int ans = -1;
	for (int i = 0; i < n; ++i) {
		ans = max(ans, calc(i));
	}
	cout << ans << endl;
}
### 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;
using ll = long long;

#define all(x) (x).begin(), (x).end()

int q[200];
vector<vector<int>> ar;

int calc(int start) {
	int beg = 0;
	int end = 1;
	q[0] = start;
	int mark[200] = {};
	mark[start] = 1;
	while (beg < end) {
		int cur = q[beg++];
		for (int to: ar[cur]) {
			if (mark[to]) {
				if (abs(mark[to] - mark[cur]) != 1) {
					return -1;
				}
				continue;
			}
			q[end++] = to;
			mark[to] = mark[cur] + 1;
		}
	}
	return mark[q[end - 1]];
}

int main() {
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	ar.resize(n);
	for (int i = 0; i < n; ++i) {
		string s;
		cin >> s;
		for (int j = i + 1; j < n; ++j) {
			if (s[j] == '1') {
				ar[i].push_back(j);
				ar[j].push_back(i);
			}
		}
	}
	int ans = -1;
	for (int i = 0; i < n; ++i) {
		ans = max(ans, calc(i));
	}
	cout << ans << endl;
}
```