#include<bits/stdc++.h>
using namespace std;

int arr[203][203] , N , col[203] , dis[203];

signed main(){
	memset(col , -1 , sizeof(col));
	cin >> N;
	for(int i = 1 ; i <= N ; ++i) for(int j = 1 ; j <= N ; ++j){char c; cin >> c; arr[i][j] = c - '0';}
	col[1] = 0; queue < int > q; q.push(1);
	while(!q.empty()){
		int t = q.front(); q.pop();
		for(int j = 1 ; j <= N ; ++j)
			if(arr[t][j])
				if(col[j] != -1 && col[j] == col[t]){puts("-1"); return 0;}
				else if(col[j] == -1){col[j] = col[t] ^ 1; q.push(j);}
	}
	int ans = 0;
	for(int i = 1 ; i <= N ; ++i){
		memset(dis , 0 , sizeof(dis)); q.push(i); dis[i] = 1; int mx = 1;
		while(!q.empty()){
			int t = q.front(); q.pop(); mx = max(mx , dis[t]);
			for(int j = 1 ; j <= N ; ++j) if(arr[t][j] && !dis[j]){dis[j] = dis[t] + 1; q.push(j);}
		}
		ans = max(ans , mx);
	}
	cout << ans; return 0;
}

### Prompt

Please formulate 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 arr[203][203] , N , col[203] , dis[203];

signed main(){
	memset(col , -1 , sizeof(col));
	cin >> N;
	for(int i = 1 ; i <= N ; ++i) for(int j = 1 ; j <= N ; ++j){char c; cin >> c; arr[i][j] = c - '0';}
	col[1] = 0; queue < int > q; q.push(1);
	while(!q.empty()){
		int t = q.front(); q.pop();
		for(int j = 1 ; j <= N ; ++j)
			if(arr[t][j])
				if(col[j] != -1 && col[j] == col[t]){puts("-1"); return 0;}
				else if(col[j] == -1){col[j] = col[t] ^ 1; q.push(j);}
	}
	int ans = 0;
	for(int i = 1 ; i <= N ; ++i){
		memset(dis , 0 , sizeof(dis)); q.push(i); dis[i] = 1; int mx = 1;
		while(!q.empty()){
			int t = q.front(); q.pop(); mx = max(mx , dis[t]);
			for(int j = 1 ; j <= N ; ++j) if(arr[t][j] && !dis[j]){dis[j] = dis[t] + 1; q.push(j);}
		}
		ans = max(ans , mx);
	}
	cout << ans; return 0;
}

```