#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <queue>

using namespace std;

int const N = 205;

vector<int> G[N];
int n;

int BFS(int S) {
	static bool vis[N];
	static int f[N];
	fill(vis + 1, vis + n + 1, 0);
	auto q = queue<int>();
	q.push(S);
	f[S] = 1;
	vis[S] = 1;
	while (!q.empty()) {
		int now = q.front();
		q.pop();
		for (auto y : G[now])
			if (vis[y] == 0) {
				f[y] = f[now] + 1;
				vis[y] = 1;
				q.push(y);
			}
	}
	int ret = *max_element(f + 1, f + n + 1);
	for (int i = 1; i <= n; ++i) 
		for (auto y : G[i])
			if (abs(f[y] - f[i]) != 1)
				ret = -1;
	return ret;
}


int main() {
	cin >> n;
	string s;
	for (int i = 1; i <= n; ++i) {
		cin >> s;
		for (int j = 1; j <= n; ++j)
			if (s[j - 1] == '1')
				G[i].push_back(j);
	}
	int ans = -1;
	for (int i = 1; i <= n; ++i) {
		auto anss = BFS(i);
		ans = max(ans, anss);
	}
	cout << ans << '\n';
}

### 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 <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <queue>

using namespace std;

int const N = 205;

vector<int> G[N];
int n;

int BFS(int S) {
	static bool vis[N];
	static int f[N];
	fill(vis + 1, vis + n + 1, 0);
	auto q = queue<int>();
	q.push(S);
	f[S] = 1;
	vis[S] = 1;
	while (!q.empty()) {
		int now = q.front();
		q.pop();
		for (auto y : G[now])
			if (vis[y] == 0) {
				f[y] = f[now] + 1;
				vis[y] = 1;
				q.push(y);
			}
	}
	int ret = *max_element(f + 1, f + n + 1);
	for (int i = 1; i <= n; ++i) 
		for (auto y : G[i])
			if (abs(f[y] - f[i]) != 1)
				ret = -1;
	return ret;
}


int main() {
	cin >> n;
	string s;
	for (int i = 1; i <= n; ++i) {
		cin >> s;
		for (int j = 1; j <= n; ++j)
			if (s[j - 1] == '1')
				G[i].push_back(j);
	}
	int ans = -1;
	for (int i = 1; i <= n; ++i) {
		auto anss = BFS(i);
		ans = max(ans, anss);
	}
	cout << ans << '\n';
}

```