#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, w[220][220], res = -1;
char p[220][220];
int main() {
	int i, j, k;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%s", p[i] + 1);
		for (j = 1; j <= n; j++) {
			if (i == j)w[i][j] = 0;
			else if (p[i][j] == '1')w[i][j] = 1;
			else w[i][j] = 1e9;
		}
	}
	for (k = 1; k <= n; k++) {
		for (i = 1; i <= n; i++) {
			for (j = 1; j <= n; j++) {
				w[i][j] = min(w[i][j], w[i][k] + w[k][j]);
			}
		}
	}
	for (i = 1; i <= n; i++) {
		int Mx = 0, chk = 0;
		for (j = 1; j <= n; j++) {
			if (Mx < w[i][j])Mx = w[i][j];
		}
		for (j = 1; j <= n; j++) {
			for (k = 1; k <= n; k++) {
				if (w[i][j] == w[i][k] && p[j][k] == '1')chk = 1;
			}
		}
		if (!chk) {
			res = max(res, Mx + 1);
		}
	}
	printf("%d\n", res);
}
### 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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, w[220][220], res = -1;
char p[220][220];
int main() {
	int i, j, k;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%s", p[i] + 1);
		for (j = 1; j <= n; j++) {
			if (i == j)w[i][j] = 0;
			else if (p[i][j] == '1')w[i][j] = 1;
			else w[i][j] = 1e9;
		}
	}
	for (k = 1; k <= n; k++) {
		for (i = 1; i <= n; i++) {
			for (j = 1; j <= n; j++) {
				w[i][j] = min(w[i][j], w[i][k] + w[k][j]);
			}
		}
	}
	for (i = 1; i <= n; i++) {
		int Mx = 0, chk = 0;
		for (j = 1; j <= n; j++) {
			if (Mx < w[i][j])Mx = w[i][j];
		}
		for (j = 1; j <= n; j++) {
			for (k = 1; k <= n; k++) {
				if (w[i][j] == w[i][k] && p[j][k] == '1')chk = 1;
			}
		}
		if (!chk) {
			res = max(res, Mx + 1);
		}
	}
	printf("%d\n", res);
}
```