#include <bits/stdc++.h>

#define F(i, a, b) for (int i = a; i <= b; i ++)
#define G(i, a, b) for (int i = a; i >= b; i --)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define mx(a, b) ((a) = max(a, b))
#define mn(a, b) ((a) = min(a, b))
#define mem(a, b) memset(a, b, sizeof a)
#define mec(a, b) memcpy(a, b, sizeof a)
#define sf scanf
#define pf printf

using namespace std;

const int N = 2e2 + 10;

char ch[N][N];
int n, vis[N], S, Ans;
int D[N], h, t;

int main() {
	sf("%d", &n);
	F(i, 1, n)
		sf("%s", ch[i] + 1);

	Ans = - 1;
	F(i, 1, n) {
		mem(vis, 0), h = 0, D[t = 1] = i, vis[i] = 1, S = - 1;
		while (h ++ < t) {
			int k = D[h], v = vis[k];
			mx(S, v);
			F(j, 1, n)
				if (ch[k][j] == '1') {
					if (!vis[j])
						D[++ t] = j, vis[j] = v + 1;
					else {
						if (vis[j] != v - 1 && vis[j] != v + 1) {
							S = - 1;
							break;
						}
					}
				}
			if (S == - 1) break;
		}
		mx(Ans, S);
	}

	pf("%d\n", Ans);
}
### 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 <bits/stdc++.h>

#define F(i, a, b) for (int i = a; i <= b; i ++)
#define G(i, a, b) for (int i = a; i >= b; i --)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define mx(a, b) ((a) = max(a, b))
#define mn(a, b) ((a) = min(a, b))
#define mem(a, b) memset(a, b, sizeof a)
#define mec(a, b) memcpy(a, b, sizeof a)
#define sf scanf
#define pf printf

using namespace std;

const int N = 2e2 + 10;

char ch[N][N];
int n, vis[N], S, Ans;
int D[N], h, t;

int main() {
	sf("%d", &n);
	F(i, 1, n)
		sf("%s", ch[i] + 1);

	Ans = - 1;
	F(i, 1, n) {
		mem(vis, 0), h = 0, D[t = 1] = i, vis[i] = 1, S = - 1;
		while (h ++ < t) {
			int k = D[h], v = vis[k];
			mx(S, v);
			F(j, 1, n)
				if (ch[k][j] == '1') {
					if (!vis[j])
						D[++ t] = j, vis[j] = v + 1;
					else {
						if (vis[j] != v - 1 && vis[j] != v + 1) {
							S = - 1;
							break;
						}
					}
				}
			if (S == - 1) break;
		}
		mx(Ans, S);
	}

	pf("%d\n", Ans);
}
```