#include<bits/stdc++.h>
using namespace std;
int n, ans, deep[220], uu;
char g[210][210];
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf("%s", g[i] + 1);
	}
	for(int u = 1; u <= n; u++){
		memset(deep, -1, sizeof(deep));
		deep[u] = 1;
		bool flag = 0;
		int res = 0;
		queue<int> q;
		q.push(u);
		while(!q.empty()){
			int x = q.front();
			q.pop();
			for(int j = 1; j <= n; j++){
				if(g[x][j] == '1'){
					if(deep[j] == deep[x] - 1 || deep[j] == deep[x] + 1) continue;
					if(deep[j] == -1){
						deep[j] = deep[x] + 1;
						res = max(res, deep[j]);
						q.push(j);
					}
					else{
						flag = 1;
						break;
					}
				}	
			}
		}
		if(flag == 0){
			if(ans < res) {
				ans = res, uu = u;
			}
		}
	}
	if(ans){
		cout<<ans;
	}
	else printf("-1");
}
### 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>
using namespace std;
int n, ans, deep[220], uu;
char g[210][210];
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf("%s", g[i] + 1);
	}
	for(int u = 1; u <= n; u++){
		memset(deep, -1, sizeof(deep));
		deep[u] = 1;
		bool flag = 0;
		int res = 0;
		queue<int> q;
		q.push(u);
		while(!q.empty()){
			int x = q.front();
			q.pop();
			for(int j = 1; j <= n; j++){
				if(g[x][j] == '1'){
					if(deep[j] == deep[x] - 1 || deep[j] == deep[x] + 1) continue;
					if(deep[j] == -1){
						deep[j] = deep[x] + 1;
						res = max(res, deep[j]);
						q.push(j);
					}
					else{
						flag = 1;
						break;
					}
				}	
			}
		}
		if(flag == 0){
			if(ans < res) {
				ans = res, uu = u;
			}
		}
	}
	if(ans){
		cout<<ans;
	}
	else printf("-1");
}
```