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

typedef long long lglg;
const int INF = 1 << 28;

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	
	int n; cin >> n;
	int adj[n][n];
	vector<pair<int,int>> edges;
	for (int i=0; i<n; ++i) {
		string x; cin >> x;
		for (int j=0; j<n; ++j) {
			if (i == j) adj[i][j] = 0;
			else if (x[j] == '1') {
				adj[i][j] = 1;
				if (i < j) edges.push_back({i, j});
			}
			else adj[i][j] = INF;
		}
	}

	for (int k=0; k<n; ++k)
		for (int i=0; i<n; ++i)
			for (int j=0; j<n; ++j)
				adj[i][j] = min(adj[i][j], adj[i][k] + adj[k][j]);
	
	bool failed = false;
	for (pair<int,int> pr : edges) {
		int x, y; tie(x, y) = pr;
		if (adj[0][x] == adj[0][y]) {
			failed = true;
			break;
		}
	}
	if (failed) cout << -1;
	else {
		int best = -1;
		for (int i=0; i<n; ++i) for (int j=0; j<n; ++j) best = max(best, adj[i][j]);
		cout << best+1;
	}
}
### Prompt

Your task is to create 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;

typedef long long lglg;
const int INF = 1 << 28;

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	
	int n; cin >> n;
	int adj[n][n];
	vector<pair<int,int>> edges;
	for (int i=0; i<n; ++i) {
		string x; cin >> x;
		for (int j=0; j<n; ++j) {
			if (i == j) adj[i][j] = 0;
			else if (x[j] == '1') {
				adj[i][j] = 1;
				if (i < j) edges.push_back({i, j});
			}
			else adj[i][j] = INF;
		}
	}

	for (int k=0; k<n; ++k)
		for (int i=0; i<n; ++i)
			for (int j=0; j<n; ++j)
				adj[i][j] = min(adj[i][j], adj[i][k] + adj[k][j]);
	
	bool failed = false;
	for (pair<int,int> pr : edges) {
		int x, y; tie(x, y) = pr;
		if (adj[0][x] == adj[0][y]) {
			failed = true;
			break;
		}
	}
	if (failed) cout << -1;
	else {
		int best = -1;
		for (int i=0; i<n; ++i) for (int j=0; j<n; ++j) best = max(best, adj[i][j]);
		cout << best+1;
	}
}
```