#include<cstdio>
#include<cstring>
#include<algorithm>
#define STB(i, s, t) for(i = s; i <= t; ++i)

using namespace std;

const int inf = 0x3f3f3f3f;
const int N = 210;
int n, ans;
int dis[N]; bool vis[N];
int mp[N][N]; 
char s[N];

void DJ(int S) {
	int i, j, u;
	memset(dis, 0x3f, sizeof dis); dis[S] = 1;
	memset(vis, 0, sizeof vis); 
	STB(i, 1, n-1) {
		u = 0;
		STB(j, 1, n)
		if(!vis[j] && dis[j] < dis[u]) u = j;
		vis[u] = 1;
		
		STB(j, 1, n)
		if(mp[u][j]) dis[j] = min(dis[j], dis[u] + 1);
	}
	
	STB(i, 1, n) STB(j, 1, n)
	if(mp[i][j] && abs(dis[i] - dis[j]) != 1) return;
	STB(i, 1, n) ans = max(ans, dis[i]);
}

int main() {
	int i, j;
	scanf("%d", &n);
	STB(i, 1, n) {
		scanf("%s", s+1);
		STB(j, 1, n) mp[i][j] = s[j] - '0';
	}
	
	ans = -1;
	STB(i, 1, n) DJ(i);
	printf("%d\n", ans);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<cstdio>
#include<cstring>
#include<algorithm>
#define STB(i, s, t) for(i = s; i <= t; ++i)

using namespace std;

const int inf = 0x3f3f3f3f;
const int N = 210;
int n, ans;
int dis[N]; bool vis[N];
int mp[N][N]; 
char s[N];

void DJ(int S) {
	int i, j, u;
	memset(dis, 0x3f, sizeof dis); dis[S] = 1;
	memset(vis, 0, sizeof vis); 
	STB(i, 1, n-1) {
		u = 0;
		STB(j, 1, n)
		if(!vis[j] && dis[j] < dis[u]) u = j;
		vis[u] = 1;
		
		STB(j, 1, n)
		if(mp[u][j]) dis[j] = min(dis[j], dis[u] + 1);
	}
	
	STB(i, 1, n) STB(j, 1, n)
	if(mp[i][j] && abs(dis[i] - dis[j]) != 1) return;
	STB(i, 1, n) ans = max(ans, dis[i]);
}

int main() {
	int i, j;
	scanf("%d", &n);
	STB(i, 1, n) {
		scanf("%s", s+1);
		STB(j, 1, n) mp[i][j] = s[j] - '0';
	}
	
	ans = -1;
	STB(i, 1, n) DJ(i);
	printf("%d\n", ans);
	return 0;
}
```