#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;

int n;
char s[205][205];
int dis[205][205];

int col[205];
bool dfs(int v, int cc)
{
	if(col[v] != 0) return col[v] == cc;
	col[v] = cc;
	rep(i, n) if(s[v][i] == '1' && !dfs(i, -cc)) return false;
	return true;
}

int main()
{
	scanf("%d", &n);
	rep(i, n) scanf("%s", s[i]);
	
	if(!dfs(0, 1)) {
		printf("-1\n");
		return 0;
	}
	
	rep(i, n) rep(j, n)
	if(i == j) dis[i][j] = 0;
	else if(s[i][j] == '1') dis[i][j] = 1;
	else dis[i][j] = INF;
	rep(k, n) rep(i, n) rep(j, n) dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
	int ans = 0;
	rep(i, n) rep(j, n) ans = max(ans, dis[i][j]);
	printf("%d\n", ans + 1);
	return 0;
}
### 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 rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;

int n;
char s[205][205];
int dis[205][205];

int col[205];
bool dfs(int v, int cc)
{
	if(col[v] != 0) return col[v] == cc;
	col[v] = cc;
	rep(i, n) if(s[v][i] == '1' && !dfs(i, -cc)) return false;
	return true;
}

int main()
{
	scanf("%d", &n);
	rep(i, n) scanf("%s", s[i]);
	
	if(!dfs(0, 1)) {
		printf("-1\n");
		return 0;
	}
	
	rep(i, n) rep(j, n)
	if(i == j) dis[i][j] = 0;
	else if(s[i][j] == '1') dis[i][j] = 1;
	else dis[i][j] = INF;
	rep(k, n) rep(i, n) rep(j, n) dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
	int ans = 0;
	rep(i, n) rep(j, n) ans = max(ans, dis[i][j]);
	printf("%d\n", ans + 1);
	return 0;
}
```