#include <bits/stdc++.h>
using namespace std;
const int N = 210;
int d[N][N], G[N][N], col[N], n; char s[N];
void dfs(int u) {
	for (int i = 1; i <= n; i++) if (G[u][i]) {
		if (!col[i]) col[i] = 3 - col[u], dfs(i);
		else if (col[u] != 3 - col[i]) {
			puts("-1"), exit(0);
		}
	}
}
int main() {
	scanf("%d", &n), memset(d, 0x3f, sizeof(d));
	for (int i = 1; i <= n; i++) {
		scanf("%s", s + 1);
		for (int j = 1; j <= n; j++) {
			G[i][j] = s[j] - '0';
			if (G[i][j] == 1) d[i][j] = 1;
		}
	}	
	col[1] = 1, dfs(1); int res = 0;
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) if (d[i][j] < n) res = max(res, d[i][j] + 1);
	printf("%d\n", res);
	
} 
### 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>
using namespace std;
const int N = 210;
int d[N][N], G[N][N], col[N], n; char s[N];
void dfs(int u) {
	for (int i = 1; i <= n; i++) if (G[u][i]) {
		if (!col[i]) col[i] = 3 - col[u], dfs(i);
		else if (col[u] != 3 - col[i]) {
			puts("-1"), exit(0);
		}
	}
}
int main() {
	scanf("%d", &n), memset(d, 0x3f, sizeof(d));
	for (int i = 1; i <= n; i++) {
		scanf("%s", s + 1);
		for (int j = 1; j <= n; j++) {
			G[i][j] = s[j] - '0';
			if (G[i][j] == 1) d[i][j] = 1;
		}
	}	
	col[1] = 1, dfs(1); int res = 0;
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) if (d[i][j] < n) res = max(res, d[i][j] + 1);
	printf("%d\n", res);
	
} 
```