#include<bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
int n, ans, u, qu[205];
char s[200][205];
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("\n%s", s[i]);
	rep(i,n){
		int de[205] = {}, b = 0, e = 0;
		de[qu[e++] = i] = 1;
		while(b < e){
			int t = qu[b++];
			rep(j,n) if(s[t][j] == '1'){
				if(de[j] == de[t]) return 0*puts("-1");
				if(!de[j]) u = de[qu[e++] = j] = de[t]+1;
			}
		}
		if(ans < u) ans = u;
	}
	printf("%d\n", ans);
}
### Prompt

In CPP, your task is to solve 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 rep(i,n) for(int i=0; i<(n); i++)
int n, ans, u, qu[205];
char s[200][205];
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("\n%s", s[i]);
	rep(i,n){
		int de[205] = {}, b = 0, e = 0;
		de[qu[e++] = i] = 1;
		while(b < e){
			int t = qu[b++];
			rep(j,n) if(s[t][j] == '1'){
				if(de[j] == de[t]) return 0*puts("-1");
				if(!de[j]) u = de[qu[e++] = j] = de[t]+1;
			}
		}
		if(ans < u) ans = u;
	}
	printf("%d\n", ans);
}
```