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

const int nm = 200;
int n;
bool g[nm][nm];
int c[nm], d[nm][nm];

bool bi(int v) {
	for (int i = 0; i < n; i++)
		if (g[v][i])
			if (c[i]) {
				if (c[i] == c[v])
					return 0;
			} else {
				c[i] = c[v] * -1;
				if (!bi(i))
					return 0;
			}
	return 1;
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		string s;
		cin >> s;
		for (int j = 0; j < n; j++) {
			g[i][j] = s[j] - '0';
			if (g[i][j])
				d[i][j] = 1;
			else if (j != i)
				d[i][j] = nm;
		}

	}

	c[0] = 1;
	if (!bi(0)) {
		cout << -1 << endl;
		return 0;
	}

	for (int k = 0; k < n; k++)
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

	int m = 0;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			m = max(m, d[i][j]);
	cout << m + 1 << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

const int nm = 200;
int n;
bool g[nm][nm];
int c[nm], d[nm][nm];

bool bi(int v) {
	for (int i = 0; i < n; i++)
		if (g[v][i])
			if (c[i]) {
				if (c[i] == c[v])
					return 0;
			} else {
				c[i] = c[v] * -1;
				if (!bi(i))
					return 0;
			}
	return 1;
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		string s;
		cin >> s;
		for (int j = 0; j < n; j++) {
			g[i][j] = s[j] - '0';
			if (g[i][j])
				d[i][j] = 1;
			else if (j != i)
				d[i][j] = nm;
		}

	}

	c[0] = 1;
	if (!bi(0)) {
		cout << -1 << endl;
		return 0;
	}

	for (int k = 0; k < n; k++)
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

	int m = 0;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			m = max(m, d[i][j]);
	cout << m + 1 << endl;
}

```