#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 200;
vector<int> g[N + 1];
int d[N + 1], n, cl[N + 1];
bool vis[N + 1];
char s[N + 2];
queue<int> q;

int bfs(int v){
	f(i, 1, n + 1)vis[i] = false;
	d[v] = 0;
	vis[v] = true;
	q.push(v);
	int an = 0;
	while (!q.empty()){
		v = q.front();
		q.pop();
		for (int u: g[v])if (!vis[u]){
			vis[u] = true;
			d[u] = d[v] + 1;
			if (d[u] > an)an = d[u];
			q.push(u);
		}
	}
	return an;
}

bool bi(int v = 1, int c = 1){
	if (cl[v])return cl[v] == c;
	cl[v] = c;
	for (int u: g[v])if (!bi(u, c ^ 3))return false;
	return true;
}

int main(){
	scanf("%d", &n);
	f(i, 1, n + 1){
		scanf("%s", s + 1);
		f(j, 1, n + 1)if (s[j] == '1')g[i].push_back(j);
	}
	if (!bi()) { printf("-1\n"); return 0; }
	int an = 0;
	f(i, 1, n + 1)an = max(an, bfs(i));
	printf("%d\n", an + 1);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 200;
vector<int> g[N + 1];
int d[N + 1], n, cl[N + 1];
bool vis[N + 1];
char s[N + 2];
queue<int> q;

int bfs(int v){
	f(i, 1, n + 1)vis[i] = false;
	d[v] = 0;
	vis[v] = true;
	q.push(v);
	int an = 0;
	while (!q.empty()){
		v = q.front();
		q.pop();
		for (int u: g[v])if (!vis[u]){
			vis[u] = true;
			d[u] = d[v] + 1;
			if (d[u] > an)an = d[u];
			q.push(u);
		}
	}
	return an;
}

bool bi(int v = 1, int c = 1){
	if (cl[v])return cl[v] == c;
	cl[v] = c;
	for (int u: g[v])if (!bi(u, c ^ 3))return false;
	return true;
}

int main(){
	scanf("%d", &n);
	f(i, 1, n + 1){
		scanf("%s", s + 1);
		f(j, 1, n + 1)if (s[j] == '1')g[i].push_back(j);
	}
	if (!bi()) { printf("-1\n"); return 0; }
	int an = 0;
	f(i, 1, n + 1)an = max(an, bfs(i));
	printf("%d\n", an + 1);
}
```