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

int main() {
	long long N, INF = 1000000000;
	vector<string> S;
	vector<vector<long long> > Dist;
	cin >> N;
	S.resize(N);
	Dist.resize(N);
	for (int i = 0; i < N; i++) {
		cin >> S[i];
		Dist[i].assign(N, INF);
		for (int j = 0; j < N; j++) {
			if (S[i][j] == '1') Dist[i][j] = 1;
		}
		Dist[i][i] = 0;
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
 				Dist[i][j] = min(Dist[i][j], Dist[i][k] + Dist[k][j]);
			}
		}
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (Dist[i][j] % 2 != (Dist[i][k] + Dist[k][j]) % 2) {
					cout << -1 << endl;
					return 0;
				}
			}
		}
	}
	long long ANS = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			ANS = max(ANS, Dist[i][j]);
		}
	}
	cout << ANS + 1 << endl;
}
### 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"
using namespace std;

int main() {
	long long N, INF = 1000000000;
	vector<string> S;
	vector<vector<long long> > Dist;
	cin >> N;
	S.resize(N);
	Dist.resize(N);
	for (int i = 0; i < N; i++) {
		cin >> S[i];
		Dist[i].assign(N, INF);
		for (int j = 0; j < N; j++) {
			if (S[i][j] == '1') Dist[i][j] = 1;
		}
		Dist[i][i] = 0;
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
 				Dist[i][j] = min(Dist[i][j], Dist[i][k] + Dist[k][j]);
			}
		}
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (Dist[i][j] % 2 != (Dist[i][k] + Dist[k][j]) % 2) {
					cout << -1 << endl;
					return 0;
				}
			}
		}
	}
	long long ANS = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			ANS = max(ANS, Dist[i][j]);
		}
	}
	cout << ANS + 1 << endl;
}
```