#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
using i64 = int64_t;
#define rep(i, j, n) for(int i = (j); i < (n); ++i)

vector<vector<int>> d;
vector<int> color;


bool dfs(int v = 0, int c = 1) {
	color[v] = c;
	rep(nv, 0, d[v].size()) {
		if (d[v][nv] == 1) {
			if (color[nv] == c) return false;
			if (color[nv] == 0 && !dfs(nv, -c)) return false;
		}
	}
	return true;
}


int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);

	int n; cin >> n;
	int ans = 0;
	char c;
	d.resize(n, vector<int>(n, 1 << 29));
	color.resize(n, 0);
	rep(i, 0, n) {
		rep(j, 0, n) {
			cin >> c;
			if (c == '1') d[i][j] = 1;
		}
		d[i][i] = 0;
	}
	if (!dfs()) {
		cout << -1;
		return 0;
	}
	
	rep(k, 0, n) rep(i, 0, n) rep(j, 0, n) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	rep(i, 0, n) rep(j, 0, n) {
		ans = max(ans, d[i][j]);
	}
	cout << ans + 1;

	return 0;
}
### 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 <iostream>
#include <algorithm>
#include <vector>
using namespace std;
using i64 = int64_t;
#define rep(i, j, n) for(int i = (j); i < (n); ++i)

vector<vector<int>> d;
vector<int> color;


bool dfs(int v = 0, int c = 1) {
	color[v] = c;
	rep(nv, 0, d[v].size()) {
		if (d[v][nv] == 1) {
			if (color[nv] == c) return false;
			if (color[nv] == 0 && !dfs(nv, -c)) return false;
		}
	}
	return true;
}


int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);

	int n; cin >> n;
	int ans = 0;
	char c;
	d.resize(n, vector<int>(n, 1 << 29));
	color.resize(n, 0);
	rep(i, 0, n) {
		rep(j, 0, n) {
			cin >> c;
			if (c == '1') d[i][j] = 1;
		}
		d[i][i] = 0;
	}
	if (!dfs()) {
		cout << -1;
		return 0;
	}
	
	rep(k, 0, n) rep(i, 0, n) rep(j, 0, n) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	rep(i, 0, n) rep(j, 0, n) {
		ans = max(ans, d[i][j]);
	}
	cout << ans + 1;

	return 0;
}
```