#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(V) (V).begin(),(V).end()

int G[201][201];
vector<ll> arr(201), arrs = arr;

int main(){
	int N;
	cin >> N;
	string S;
	for (int i = 0;i < N;i++) {
		cin >> S;
		for (int j = 0;j < N;j++) {
			if (S[j] == '0')G[i][j] = 0;
			else G[i][j] = 1;
		}
	}
	ll cnt = 0;
	ll a, x = 0, mx = 0;
	queue<int> Q, QS;
	QS.push(-1);
	bool bo = 0;
	for (int i = 0;i < N;i++) {
		bo = 0;
		x = 0;
		arr = arrs;
		Q = QS;
		Q.push(i);
		while (Q.size() > 1) {
			a = Q.front();
			Q.pop();
			if (a < 0) {
				x++;
				Q.push(-1);
				continue;
			}
			if (arr[a] != 0)continue;
			arr[a] = x;
			for (int j = 0;j < N;j++) {
				if (G[a][j]) {
					if (arr[j] != 0){
						if (abs(arr[j] - x) != 1) {
							bo = 1;break;
						}
						else
							continue;
					}
					Q.push(j);
				}
			}
			if (bo)break;
		}
		if (!bo) {
			mx = max(mx, x);
		}
	}
	if (mx != 0)cout << mx << endl;
	else cout << -1 << endl;
}
### Prompt

Your challenge is to write 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>
using namespace std;
#define ll long long
#define all(V) (V).begin(),(V).end()

int G[201][201];
vector<ll> arr(201), arrs = arr;

int main(){
	int N;
	cin >> N;
	string S;
	for (int i = 0;i < N;i++) {
		cin >> S;
		for (int j = 0;j < N;j++) {
			if (S[j] == '0')G[i][j] = 0;
			else G[i][j] = 1;
		}
	}
	ll cnt = 0;
	ll a, x = 0, mx = 0;
	queue<int> Q, QS;
	QS.push(-1);
	bool bo = 0;
	for (int i = 0;i < N;i++) {
		bo = 0;
		x = 0;
		arr = arrs;
		Q = QS;
		Q.push(i);
		while (Q.size() > 1) {
			a = Q.front();
			Q.pop();
			if (a < 0) {
				x++;
				Q.push(-1);
				continue;
			}
			if (arr[a] != 0)continue;
			arr[a] = x;
			for (int j = 0;j < N;j++) {
				if (G[a][j]) {
					if (arr[j] != 0){
						if (abs(arr[j] - x) != 1) {
							bo = 1;break;
						}
						else
							continue;
					}
					Q.push(j);
				}
			}
			if (bo)break;
		}
		if (!bo) {
			mx = max(mx, x);
		}
	}
	if (mx != 0)cout << mx << endl;
	else cout << -1 << endl;
}
```