#include<bits/stdc++.h>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 203;
int n, dis[N], ans, q[N], front, rear;
bool s[N][N];
inline void bfs(int x){
	memset(dis, -1, sizeof dis);
	front = rear = dis[x] = 0; q[rear ++] = x;
	while(front < rear){
		int now = q[front ++];
		for(Rint i = 1;i <= n;i ++)
			if(s[now][i]){
				if(~dis[i]){
					if((dis[i] & 1) == (dis[now] & 1)){puts("-1"); exit(0);}
				} else {
					dis[i] = dis[now] + 1;
					q[rear ++] = i;
				}
			}
	}
}
int main(){
	scanf("%d", &n);
	for(Rint i = 1;i <= n;i ++)
		for(Rint j = 1;j <= n;j ++){
			int ch = getchar();
			while(ch != '0' && ch != '1') ch = getchar();
			s[i][j] = (ch == '1');
		}
	for(Rint i = 1;i <= n;i ++){
		bfs(i);
		for(Rint j = 1;j <= n;j ++) ans = max(ans, dis[j]);
	}
	printf("%d", ans + 1);
}
### Prompt

Generate 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 Rint register int
using namespace std;
typedef long long LL;
const int N = 203;
int n, dis[N], ans, q[N], front, rear;
bool s[N][N];
inline void bfs(int x){
	memset(dis, -1, sizeof dis);
	front = rear = dis[x] = 0; q[rear ++] = x;
	while(front < rear){
		int now = q[front ++];
		for(Rint i = 1;i <= n;i ++)
			if(s[now][i]){
				if(~dis[i]){
					if((dis[i] & 1) == (dis[now] & 1)){puts("-1"); exit(0);}
				} else {
					dis[i] = dis[now] + 1;
					q[rear ++] = i;
				}
			}
	}
}
int main(){
	scanf("%d", &n);
	for(Rint i = 1;i <= n;i ++)
		for(Rint j = 1;j <= n;j ++){
			int ch = getchar();
			while(ch != '0' && ch != '1') ch = getchar();
			s[i][j] = (ch == '1');
		}
	for(Rint i = 1;i <= n;i ++){
		bfs(i);
		for(Rint j = 1;j <= n;j ++) ans = max(ans, dis[j]);
	}
	printf("%d", ans + 1);
}
```