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

const int MAXN = 205;
const int INF = 1 << 29;

int N;
char S[MAXN][MAXN];
int dis[MAXN][MAXN];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	cin >> N;
	for(int i = 0; i < N; i++) cin >> S[i];

	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) dis[i][j] = i == j ? 0 : INF;
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++)
		if(S[i][j] == '1') dis[i][j] = 1;
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) for(int k = 0; k < N; k++)
		dis[j][k] = min(dis[j][k], dis[j][i] + dis[i][k]);
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++)
		if(S[i][j] == '1') if(dis[0][i] % 2 == dis[0][j] % 2) {
			cout << "-1\n";
			return 0;
		}
	int mx = 0;
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) mx = max(mx, dis[i][j]);
	cout << mx + 1 << "\n";
	return 0;
}

### 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;

const int MAXN = 205;
const int INF = 1 << 29;

int N;
char S[MAXN][MAXN];
int dis[MAXN][MAXN];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	cin >> N;
	for(int i = 0; i < N; i++) cin >> S[i];

	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) dis[i][j] = i == j ? 0 : INF;
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++)
		if(S[i][j] == '1') dis[i][j] = 1;
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) for(int k = 0; k < N; k++)
		dis[j][k] = min(dis[j][k], dis[j][i] + dis[i][k]);
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++)
		if(S[i][j] == '1') if(dis[0][i] % 2 == dis[0][j] % 2) {
			cout << "-1\n";
			return 0;
		}
	int mx = 0;
	for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) mx = max(mx, dis[i][j]);
	cout << mx + 1 << "\n";
	return 0;
}

```