#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cmath>
#include <tuple>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int n;
string s[200];

signed main() {
	int i, j, k;
	
	cin >> n;
	rep(i, n) cin >> s[i];
	
	int ans = -1;
	rep(i, n) {
		queue<int> que;
		int dist[200];
		rep(j, n) dist[j] = -1;
		
		que.push(i);
		dist[i] = 0;
		while (!que.empty()) {
			int v = que.front(); que.pop();
			rep(j, n) {
				if (s[v][j] == '1' && dist[j] == -1) {
					dist[j] = dist[v] + 1;
					que.push(j);
				}
			}
		}
		
		rep(j, n) {
			for (k = j + 1; k < n; k++) {
				if (s[j][k] == '0') continue;
				if (abs(dist[j] - dist[k]) != 1) break;
			}
			if (k < n) break;
		}
		if (j == n) {
			int res = 0;
			rep(k, n) res = max(res, dist[k]);
			ans = max(ans, res + 1);
		}
	}
	cout << ans << endl;
	return 0;
}
### 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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cmath>
#include <tuple>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int n;
string s[200];

signed main() {
	int i, j, k;
	
	cin >> n;
	rep(i, n) cin >> s[i];
	
	int ans = -1;
	rep(i, n) {
		queue<int> que;
		int dist[200];
		rep(j, n) dist[j] = -1;
		
		que.push(i);
		dist[i] = 0;
		while (!que.empty()) {
			int v = que.front(); que.pop();
			rep(j, n) {
				if (s[v][j] == '1' && dist[j] == -1) {
					dist[j] = dist[v] + 1;
					que.push(j);
				}
			}
		}
		
		rep(j, n) {
			for (k = j + 1; k < n; k++) {
				if (s[j][k] == '0') continue;
				if (abs(dist[j] - dist[k]) != 1) break;
			}
			if (k < n) break;
		}
		if (j == n) {
			int res = 0;
			rep(k, n) res = max(res, dist[k]);
			ans = max(ans, res + 1);
		}
	}
	cout << ans << endl;
	return 0;
}
```