#include <bits/stdc++.h>
using namespace std;
using Int = int_fast64_t;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	Int n; cin >> n;
	vector<string> s(n);
	for(auto &i:s) cin >> i;
	bool bpg = true;
	vector<Int> c(n, -1);
	queue<Int> q;
	c[0] = 0;
	q.push(0);
	while(!q.empty()){
		Int p = q.front(); q.pop();
		for(Int i=0; i<n; ++i)
			if(s[p][i] == '1'){
				if(c[i] != -1 && c[i] != 1-c[p]) bpg = false;
				if(c[i] == -1) q.push(i);
				c[i] = 1-c[p];
			}
	}
	if(!bpg){
		cout << -1 << "\n";
		return 0;
	}
	vector<vector<Int>> d(n, vector<Int>(n, 201));
	for(Int i=0; i<n; ++i)
		for(Int j=0; j<n; ++j){
			if(i == j) d[i][j] = 0;
			else if(s[i][j] == '1') d[i][j] = 1;
		}
	for(Int k=0; k<n; ++k)
		for(Int i=0; i<n; ++i)
			for(Int j=0; j<n; ++j)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	Int ans = 2;
	for(Int i=0; i<n; ++i){
		Int tmp = 0;
		for(Int j=0; j<n; ++j) tmp = max(tmp, d[i][j]);
		ans = max(ans, tmp+1);
	}
	cout << ans << "\n";
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;
using Int = int_fast64_t;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	Int n; cin >> n;
	vector<string> s(n);
	for(auto &i:s) cin >> i;
	bool bpg = true;
	vector<Int> c(n, -1);
	queue<Int> q;
	c[0] = 0;
	q.push(0);
	while(!q.empty()){
		Int p = q.front(); q.pop();
		for(Int i=0; i<n; ++i)
			if(s[p][i] == '1'){
				if(c[i] != -1 && c[i] != 1-c[p]) bpg = false;
				if(c[i] == -1) q.push(i);
				c[i] = 1-c[p];
			}
	}
	if(!bpg){
		cout << -1 << "\n";
		return 0;
	}
	vector<vector<Int>> d(n, vector<Int>(n, 201));
	for(Int i=0; i<n; ++i)
		for(Int j=0; j<n; ++j){
			if(i == j) d[i][j] = 0;
			else if(s[i][j] == '1') d[i][j] = 1;
		}
	for(Int k=0; k<n; ++k)
		for(Int i=0; i<n; ++i)
			for(Int j=0; j<n; ++j)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	Int ans = 2;
	for(Int i=0; i<n; ++i){
		Int tmp = 0;
		for(Int j=0; j<n; ++j) tmp = max(tmp, d[i][j]);
		ans = max(ans, tmp+1);
	}
	cout << ans << "\n";
}
```