#include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, res, vis[200];
bool b[200][200];


int main() {
	cin >> N;
	res = -1;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			char c;
			cin >> c;
			b[i][j] = (c == '1');
		}
	}
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			vis[j] = 0;
		}
		int c = 0;
		queue<int>q;
		q.push(i);
		bool f = false;
		while (!q.empty()) {
			queue<int>r;
			++c;
			while (!q.empty()) {
				int x = q.front();
				q.pop();
				if (vis[x] == c)continue;
				if (vis[x] != 0) {
					f = true;
					break;
				}
				else {
					vis[x] = c;
					for (int j = 0; j < N; ++j) {
						if (b[x][j] && !vis[j]) {
							r.push(j);
						}
					}
				}
			}
			if (f)break;
			q = r;
		}
		if (!f) {
			res = max(res, c);
		}
	}
	cout << res << endl;
}
### Prompt

Generate 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>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, res, vis[200];
bool b[200][200];


int main() {
	cin >> N;
	res = -1;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			char c;
			cin >> c;
			b[i][j] = (c == '1');
		}
	}
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			vis[j] = 0;
		}
		int c = 0;
		queue<int>q;
		q.push(i);
		bool f = false;
		while (!q.empty()) {
			queue<int>r;
			++c;
			while (!q.empty()) {
				int x = q.front();
				q.pop();
				if (vis[x] == c)continue;
				if (vis[x] != 0) {
					f = true;
					break;
				}
				else {
					vis[x] = c;
					for (int j = 0; j < N; ++j) {
						if (b[x][j] && !vis[j]) {
							r.push(j);
						}
					}
				}
			}
			if (f)break;
			q = r;
		}
		if (!f) {
			res = max(res, c);
		}
	}
	cout << res << endl;
}
```