#include <bits/stdc++.h>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}
int main() {
	int n = ri();
	std::string s[n];
	for (int i = 0; i < n; i++) std::cin >> s[i];
	
	int res = -1;
	for (int i = 0; i < n; i++) {
		std::queue<int> que;
		que.push(i);
		std::vector<int> dist(n, -1);
		dist[i] = 0;
		bool can = true;
		while (que.size()) {
			int i = que.front();
			que.pop();
			for (int j = 0; j < n; j++) {
				if (s[i][j] == '1') {
					if (dist[j] != -1) {
						if ((dist[j] & 1) == (dist[i] & 1)) {
							can = false;
							break;
						}
					} else {
						dist[j] = dist[i] + 1;
						que.push(j);
					}
				}
			}
			if (!can) break;
		}
		if (can) res = std::max(res, 1 + *std::max_element(dist.begin(), dist.end()));
	}
	std::cout << res << std::endl;
	
	return 0;
}

### Prompt

Please formulate 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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}
int main() {
	int n = ri();
	std::string s[n];
	for (int i = 0; i < n; i++) std::cin >> s[i];
	
	int res = -1;
	for (int i = 0; i < n; i++) {
		std::queue<int> que;
		que.push(i);
		std::vector<int> dist(n, -1);
		dist[i] = 0;
		bool can = true;
		while (que.size()) {
			int i = que.front();
			que.pop();
			for (int j = 0; j < n; j++) {
				if (s[i][j] == '1') {
					if (dist[j] != -1) {
						if ((dist[j] & 1) == (dist[i] & 1)) {
							can = false;
							break;
						}
					} else {
						dist[j] = dist[i] + 1;
						que.push(j);
					}
				}
			}
			if (!can) break;
		}
		if (can) res = std::max(res, 1 + *std::max_element(dist.begin(), dist.end()));
	}
	std::cout << res << std::endl;
	
	return 0;
}

```