#include <bits/stdc++.h>

#define PI 3.14159265358979323846264338327950288419

using namespace std;

int n, layer[202];
int adj[202][202];
char eat[202];

int ans = -1;

int que[202], l, r;

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf(" %s", eat + 1);
		for(int j = 1; j <= n; j++){
			adj[i][j] = eat[j] - '0';
		}
	}
	for(int i = 1; i <= n; i++){
		memset(layer, -1, sizeof(layer));
		l = 0, r = 1, que[1] = i;
		layer[i] = 0;
		
		int max_layer = 0;
		
		while(l < r){
			int cur = que[++l];
			for(int j = 1; j <= n; j++){
				if(adj[cur][j] == 1 && layer[j] == -1){
					layer[j] = layer[cur] + 1;
					que[++r] = j;
					max_layer = layer[j];
				}
			}
		}
		
		bool ok = true;
		
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= n; j++){
				if(adj[i][j] && abs(layer[i] - layer[j]) != 1){
					ok = false;
					break;
				}
			}
		}
		
		if(ok){
			ans = max(ans, max_layer + 1);
		}
	}
	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 PI 3.14159265358979323846264338327950288419

using namespace std;

int n, layer[202];
int adj[202][202];
char eat[202];

int ans = -1;

int que[202], l, r;

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf(" %s", eat + 1);
		for(int j = 1; j <= n; j++){
			adj[i][j] = eat[j] - '0';
		}
	}
	for(int i = 1; i <= n; i++){
		memset(layer, -1, sizeof(layer));
		l = 0, r = 1, que[1] = i;
		layer[i] = 0;
		
		int max_layer = 0;
		
		while(l < r){
			int cur = que[++l];
			for(int j = 1; j <= n; j++){
				if(adj[cur][j] == 1 && layer[j] == -1){
					layer[j] = layer[cur] + 1;
					que[++r] = j;
					max_layer = layer[j];
				}
			}
		}
		
		bool ok = true;
		
		for(int i = 1; i <= n; i++){
			for(int j = 1; j <= n; j++){
				if(adj[i][j] && abs(layer[i] - layer[j]) != 1){
					ok = false;
					break;
				}
			}
		}
		
		if(ok){
			ans = max(ans, max_layer + 1);
		}
	}
	printf("%d\n", ans);
}
```