#include<bits/stdc++.h>
using namespace std;


#define rep(i, n) for(int i = 0; i < (n); i++)
using ULL = unsigned long long;

int n;

using node = bool[200];
void input(node &d) {
	rep(i, n) {
		char c; cin >> c;
		d[i] = c == '1';
	}
}

node v[200];
bool p[200];

vector<vector<int>> e;

void solve() {
	cin >> n;
	rep(i, n) { input(v[i]); }

	e.resize(n);
	rep(i, n) {
		rep(j, n) {
			if (v[i][j])e[i].push_back(j);
		}
	}

	int ans = 0;
	rep(i, n) {
		int d[200]; rep(j, n) d[j] = 99999;
		queue<int> q;
		q.push(i); d[i] = 1;
		while (!q.empty()) {
			int p = q.front(); q.pop();
			ans = max(ans, d[p]);
			for (int pp : e[p]) {
				if (d[pp] != 99999) {
					if ((d[p] - d[pp]) % 2 == 0) {
						cout << -1 << endl;
						return;
					}
					continue;
				}
				d[pp] = d[p] + 1;
				q.push(pp);
			}
		}
	}

	if (ans == 1) cout << -1 << endl;
	 else cout << ans << endl;
}

int main() {

	solve();

	return 0;
}
### 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;


#define rep(i, n) for(int i = 0; i < (n); i++)
using ULL = unsigned long long;

int n;

using node = bool[200];
void input(node &d) {
	rep(i, n) {
		char c; cin >> c;
		d[i] = c == '1';
	}
}

node v[200];
bool p[200];

vector<vector<int>> e;

void solve() {
	cin >> n;
	rep(i, n) { input(v[i]); }

	e.resize(n);
	rep(i, n) {
		rep(j, n) {
			if (v[i][j])e[i].push_back(j);
		}
	}

	int ans = 0;
	rep(i, n) {
		int d[200]; rep(j, n) d[j] = 99999;
		queue<int> q;
		q.push(i); d[i] = 1;
		while (!q.empty()) {
			int p = q.front(); q.pop();
			ans = max(ans, d[p]);
			for (int pp : e[p]) {
				if (d[pp] != 99999) {
					if ((d[p] - d[pp]) % 2 == 0) {
						cout << -1 << endl;
						return;
					}
					continue;
				}
				d[pp] = d[p] + 1;
				q.push(pp);
			}
		}
	}

	if (ans == 1) cout << -1 << endl;
	 else cout << ans << endl;
}

int main() {

	solve();

	return 0;
}
```